Motion in a Plane Test 8

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Motion in a Plane Test 8

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Question 1
1 / -0

Basketball and projectile
Using the laws of physics, it is possible to make a successful jump shot in basketball every time. The trajectory of a basketball is always is a parabola when it is launched up into the air, and this is due to the affects of gravity as well as the force put on the ball by the player and its property follows the properties of a projectile.
The ball's velocity changes as it moves through the air, but the velocity of the first half of the path matches with the last half of the path. It may deviate from its path if the ball collides with either the backboard or another player interfering with its path. From the information of the height of the player, the horizontal distance and height of the hoop, the player can easily calculate the speed to be imparted to the ball and the angle for a sure success. If player’s height is 1.27 meters standing at a distance of 2 meters from the hoop which is at an height of 3.05 meters, he needs an angle of 55° and velocity of the ball 7 m/s to be imparted to make his shot a success. (If he uses a smaller angle has to shoot with a greater velocity.) If he stands 6 meters distance from the hoop, he needs an angle 60° and a velocity 9.5 m/s to be successful. To improve chances of an accurate shot, often a player includes a backspin on the ball as he launches it for a shot. The backspin ensures that the ball enters the hoop, especially if the shot is a "soft shot." Soft shot is when the ball is shot at a low angle and low velocity, the player adds a backspin because if the ball winds up hitting the rim, the spin will help it to enter the hoop. The backspin changes the velocity direction (once it hits the rim) to the opposite direction of the rim rather than bouncing it out.
The trajectory of a basketball is always is a

A
A. Straight line

B
B. Circle

C
C. Parabola

D
D. May be a circle, may be a parabola

Solution

The trajectory of a basketball is always is a parabola when it is launched up into the air, and this is due to the affects of gravity as well as the force put on the ball by the player and its property follows the properties of a projectile. The ball's velocity changes as it moves through the air, but the velocity of the first half of the path matches with the last half of the path.
Question 2
1 / -0

Basketball and projectile
Using the laws of physics, it is possible to make a successful jump shot in basketball every time. The trajectory of a basketball is always is a parabola when it is launched up into the air, and this is due to the affects of gravity as well as the force put on the ball by the player and its property follows the properties of a projectile.
The ball's velocity changes as it moves through the air, but the velocity of the first half of the path matches with the last half of the path. It may deviate from its path if the ball collides with either the backboard or another player interfering with its path. From the information of the height of the player, the horizontal distance and height of the hoop, the player can easily calculate the speed to be imparted to the ball and the angle for a sure success. If player’s height is 1.27 meters standing at a distance of 2 meters from the hoop which is at an height of 3.05 meters, he needs an angle of 55° and velocity of the ball 7 m/s to be imparted to make his shot a success. (If he uses a smaller angle has to shoot with a greater velocity.) If he stands 6 meters distance from the hoop, he needs an angle 60° and a velocity 9.5 m/s to be successful. To improve chances of an accurate shot, often a player includes a backspin on the ball as he launches it for a shot. The backspin ensures that the ball enters the hoop, especially if the shot is a "soft shot." Soft shot is when the ball is shot at a low angle and low velocity, the player adds a backspin because if the ball winds up hitting the rim, the spin will help it to enter the hoop. The backspin changes the velocity direction (once it hits the rim) to the opposite direction of the rim rather than bouncing it out.
To calculate the speed to be imparted to the ball and the angle for an accurate shot the required information are

A
A. Height of the player

B
B. Horizontal distance of the hoop

C
C. Height of the hoop

D
D. All of the above

Solution

To calculate the speed to be imparted to the ball and the angle for an accurate shot the required information are height of the player, Horizontal distance of the hoop and Height of the hoop.
Question 3
1 / -0

Basketball and projectile
Using the laws of physics, it is possible to make a successful jump shot in basketball every time. The trajectory of a basketball is always is a parabola when it is launched up into the air, and this is due to the affects of gravity as well as the force put on the ball by the player and its property follows the properties of a projectile.
The ball's velocity changes as it moves through the air, but the velocity of the first half of the path matches with the last half of the path. It may deviate from its path if the ball collides with either the backboard or another player interfering with its path. From the information of the height of the player, the horizontal distance and height of the hoop, the player can easily calculate the speed to be imparted to the ball and the angle for a sure success. If player’s height is 1.27 meters standing at a distance of 2 meters from the hoop which is at an height of 3.05 meters, he needs an angle of 55° and velocity of the ball 7 m/s to be imparted to make his shot a success. (If he uses a smaller angle has to shoot with a greater velocity.) If he stands 6 meters distance from the hoop, he needs an angle 60° and a velocity 9.5 m/s to be successful. To improve chances of an accurate shot, often a player includes a backspin on the ball as he launches it for a shot. The backspin ensures that the ball enters the hoop, especially if the shot is a "soft shot." Soft shot is when the ball is shot at a low angle and low velocity, the player adds a backspin because if the ball winds up hitting the rim, the spin will help it to enter the hoop. The backspin changes the velocity direction (once it hits the rim) to the opposite direction of the rim rather than bouncing it out.
Why backspin applied on basketball while launching it for a shot?

A
A. The backspin ensures that the ball follows parabolic path

B
B. The backspin ensures that the ball enters the hoop

C
C. The backspin ensures that the ball does not collide with an interfering player

D
D. The backspin ensures that the ball moves faster

Solution

To improve chances of an accurate shot, often a player includes a backspin on the ball as he launches it for a shot. The backspin ensures that the ball enters the hoop, especially if the shot is a "soft shot." Soft shot is when the ball is shot at a low angle and low velocity, the player adds a backspin because if the ball winds up hitting the rim, the spin will help it to enter the hoop. The backspin changes the velocity direction (once it hits the rim) to the opposite direction of the rim rather than bouncing it out.
Question 4
1 / -0

Projectile motion to orbital motion:
If we throw a ball horizontally with a speed 8000 m/s off the surface of the Earth (and there are no obstructions in the way) how far will it travel in the vertical and horizontal directions in 1 second? (Ignore the air resistance)
v_x = 8000 m/s
t = 1s
So, horizontally it will move
v_yt = 8000 m
V_y = 0
t = 1s
g = 10 m/s²
So, vertically it will move down
1/2 gt² = 5 m
But the curvature of the Earth changes by 5m in every 8000 m. So, will the ball ever touch the earth? The answer is NO! The ball will continually falls in free fall around the Earth and ultimately it will be orbiting around the Earth!
If v = 8000 m/s the orbit will be circular.
If v is within the range 8000 m/s to 11200 m/s, then the orbit will be elliptical. 11200 m/s is the velocity required to escape the earth. Escape velocity is the velocity of the projectile body that is sufficient for it to escape the gravitational attraction of earth. 42500 m/s is the velocity required to escape the Solar System!
What is the upper limit of the speed for a particle thrown horizontally so that it remains as a projectile?

A
A. v > 8000 m/s

B
B. v < 8000 m/s

C
C. There is no such limit.

D
D. The limit depends on the weight of the particle.

Solution

If the speed is 8000 m/s, in 1 second, it moves horizontally 8000 m and vertically 5m. But the curvature of the Earth changes by 5m in every 8000 m. So, it will not touch the earth. So, to remain a projectile, the speed should be less than 8000 m/s.
Question 5
1 / -0

Projectile motion to orbital motion:
If we throw a ball horizontally with a speed 8000 m/s off the surface of the Earth (and there are no obstructions in the way) how far will it travel in the vertical and horizontal directions in 1 second? (Ignore the air resistance)
v_x = 8000 m/s
t = 1s
So, horizontally it will move
v_yt = 8000 m
V_y = 0
t = 1s
g = 10 m/s²
So, vertically it will move down
1/2 gt² = 5 m
But the curvature of the Earth changes by 5m in every 8000 m. So, will the ball ever touch the earth? The answer is NO! The ball will continually falls in free fall around the Earth and ultimately it will be orbiting around the Earth!
If v = 8000 m/s the orbit will be circular.
If v is within the range 8000 m/s to 11200 m/s, then the orbit will be elliptical. 11200 m/s is the velocity required to escape the earth. Escape velocity is the velocity of the projectile body that is sufficient for it to escape the gravitational attraction of earth. 42500 m/s is the velocity required to escape the Solar System!
When a particle orbits the earth in a circular or elliptical path then the particle

A
A. is beyond the gravitational attraction

B
B. is under the action of gravitational attraction

C
C. becomes called a satellite

D
D. Both (a) and (c)

Solution

Gravity supplies the necessary centripetal force to hold a satellite in orbit.
Question 6
1 / -0

Projectile motion to orbital motion:
If we throw a ball horizontally with a speed 8000 m/s off the surface of the Earth (and there are no obstructions in the way) how far will it travel in the vertical and horizontal directions in 1 second? (Ignore the air resistance)
v_x = 8000 m/s
t = 1s
So, horizontally it will move
v_yt = 8000 m
V_y = 0
t = 1s
g = 10 m/s²
So, vertically it will move down
1/2 gt² = 5 m
But the curvature of the Earth changes by 5m in every 8000 m. So, will the ball ever touch the earth? The answer is NO! The ball will continually falls in free fall around the Earth and ultimately it will be orbiting around the Earth!
If v = 8000 m/s the orbit will be circular.
If v is within the range 8000 m/s to 11200 m/s, then the orbit will be elliptical. 11200 m/s is the velocity required to escape the earth. Escape velocity is the velocity of the projectile body that is sufficient for it to escape the gravitational attraction of earth. 42500 m/s is the velocity required to escape the Solar System!
Escape velocity from earth is ............... escape velocity from solar system

A
A. Greater than

B
B. Smaller than

C
C. Equal to

D
D. Greater than or equal to

Solution

Escape velocity from earth is 11200 m/s and that from solar system is 42500 m/s.
Question 7
1 / -0

Projectile motion to orbital motion:
If we throw a ball horizontally with a speed 8000 m/s off the surface of the Earth (and there are no obstructions in the way) how far will it travel in the vertical and horizontal directions in 1 second? (Ignore the air resistance)
v_x = 8000 m/s
t = 1s
So, horizontally it will move
v_yt = 8000 m
V_y = 0
t = 1s
g = 10 m/s²
So, vertically it will move down
1/2 gt² = 5 m
But the curvature of the Earth changes by 5m in every 8000 m. So, will the ball ever touch the earth? The answer is NO! The ball will continually falls in free fall around the Earth and ultimately it will be orbiting around the Earth!
If v = 8000 m/s the orbit will be circular.
If v is within the range 8000 m/s to 11200 m/s, then the orbit will be elliptical. 11200 m/s is the velocity required to escape the earth. Escape velocity is the velocity of the projectile body that is sufficient for it to escape the gravitational attraction of earth. 42500 m/s is the velocity required to escape the Solar System!
What is the value of escape velocity for earth?

A
A. 8000 m/s

B
B. 11200 m/s

C
C. Less than 8000 m/s

D
D. None of the above

Solution

11200 m/s is the escape velocity of the earth. If a particle is projected with this velocity it escapes the gravitational attraction of earth.
Question 8
1 / -0

Projectile motion to orbital motion:
If we throw a ball horizontally with a speed 8000 m/s off the surface of the Earth (and there are no obstructions in the way) how far will it travel in the vertical and horizontal directions in 1 second? (Ignore the air resistance)
v_x = 8000 m/s
t = 1s
So, horizontally it will move
v_yt = 8000 m
V_y = 0
t = 1s
g = 10 m/s²
So, vertically it will move down
1/2 gt² = 5 m
But the curvature of the Earth changes by 5m in every 8000 m. So, will the ball ever touch the earth? The answer is NO! The ball will continually falls in free fall around the Earth and ultimately it will be orbiting around the Earth!
If v = 8000 m/s the orbit will be circular.
If v is within the range 8000 m/s to 11200 m/s, then the orbit will be elliptical. 11200 m/s is the velocity required to escape the earth. Escape velocity is the velocity of the projectile body that is sufficient for it to escape the gravitational attraction of earth. 42500 m/s is the velocity required to escape the Solar System!
Curvature of the Earth changes by………... in every

A
A. 5m, 8000 km

B
B. 0.005 km, 8 km

C
C. 5m, 8 km

D
D. 5m, 80000 m

Solution

curvature of the Earth changes by 5m in every 8000 m i.e. 0.005 km in every 8 km.
Question 9
1 / -0

Basketball and projectile
Using the laws of physics, it is possible to make a successful jump shot in basketball every time. The trajectory of a basketball is always is a parabola when it is launched up into the air, and this is due to the affects of gravity as well as the force put on the ball by the player and its property follows the properties of a projectile.
The ball's velocity changes as it moves through the air, but the velocity of the first half of the path matches with the last half of the path. It may deviate from its path if the ball collides with either the backboard or another player interfering with its path. From the information of the height of the player, the horizontal distance and height of the hoop, the player can easily calculate the speed to be imparted to the ball and the angle for a sure success. If player’s height is 1.27 meters standing at a distance of 2 meters from the hoop which is at an height of 3.05 meters, he needs an angle of 55° and velocity of the ball 7 m/s to be imparted to make his shot a success. (If he uses a smaller angle has to shoot with a greater velocity.) If he stands 6 meters distance from the hoop, he needs an angle 60° and a velocity 9.5 m/s to be successful. To improve chances of an accurate shot, often a player includes a backspin on the ball as he launches it for a shot. The backspin ensures that the ball enters the hoop, especially if the shot is a "soft shot." Soft shot is when the ball is shot at a low angle and low velocity, the player adds a backspin because if the ball winds up hitting the rim, the spin will help it to enter the hoop. The backspin changes the velocity direction (once it hits the rim) to the opposite direction of the rim rather than bouncing it out.
When the basketball deviates from its parabolic path?

A
A. if the ball collides with the backboard

B
B. if the ball collide with another player interfering with its path

C
C. if a backspin is applied to the ball

D
D. Both (a) and (b)

Solution

The basketball may deviate from its path if the ball collides with either the backboard or another player interfering with its path. Backspin does not deviate the ball from its trajectory. It improves the chances of an accurate shot.
Question 10
1 / -0

Basketball and projectile
Using the laws of physics, it is possible to make a successful jump shot in basketball every time. The trajectory of a basketball is always is a parabola when it is launched up into the air, and this is due to the affects of gravity as well as the force put on the ball by the player and its property follows the properties of a projectile.
The ball's velocity changes as it moves through the air, but the velocity of the first half of the path matches with the last half of the path. It may deviate from its path if the ball collides with either the backboard or another player interfering with its path. From the information of the height of the player, the horizontal distance and height of the hoop, the player can easily calculate the speed to be imparted to the ball and the angle for a sure success. If player’s height is 1.27 meters standing at a distance of 2 meters from the hoop which is at an height of 3.05 meters, he needs an angle of 55° and velocity of the ball 7 m/s to be imparted to make his shot a success. (If he uses a smaller angle has to shoot with a greater velocity.) If he stands 6 meters distance from the hoop, he needs an angle 60° and a velocity 9.5 m/s to be successful. To improve chances of an accurate shot, often a player includes a backspin on the ball as he launches it for a shot. The backspin ensures that the ball enters the hoop, especially if the shot is a "soft shot." Soft shot is when the ball is shot at a low angle and low velocity, the player adds a backspin because if the ball winds up hitting the rim, the spin will help it to enter the hoop. The backspin changes the velocity direction (once it hits the rim) to the opposite direction of the rim rather than bouncing it out.
Why backspin applied on basketball while launching it for a shot?

A
A. The backspin ensures that the ball follows parabolic path

B
B. The backspin ensures that the ball enters the hoop

C
C. The backspin ensures that the ball does not collide with an interfering player

D
D. The backspin ensures that the ball moves faster

Solution

To improve chances of an accurate shot, often a player includes a backspin on the ball as he launches it for a shot. The backspin ensures that the ball enters the hoop, especially if the shot is a "soft shot." Soft shot is when the ball is shot at a low angle and low velocity, the player adds a backspin because if the ball winds up hitting the rim, the spin will help it to enter the hoop. The backspin changes the velocity direction (once it hits the rim) to the opposite direction of the rim rather than bouncing it out.