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Motion in a Plane Test - 7

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Motion in a Plane Test - 7
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  • Question 1
    4 / -1

     

    Which of the following is true regarding projectile motion?

     

    Solution

     

     

    Horizontal velocity of a projectile is not affected by gravity.

     

     

  • Question 2
    4 / -1

     

    A bomb is released by a horizontal flying aeroplane. The trajectory of the bomb is

     

    Solution

     

     

    It is an example of projectile motion. Therefore, the trajectory of the bomb is parabola.

     

     

  • Question 3
    4 / -1

     

     In case of a projectile motion, what is the angle between the velocity and acceleration at the highest point?

     

    Solution

     

     

    At the highest point, velocity is acting horizontally and acceleration ( = acceleration due to gravity) is acting vertically downwards. Therefore, at the highest point the angle between velocity and acceleration is 90∘ .

     

     

  • Question 4
    4 / -1

     

    If a body is projected with an angle θ to the horizontal, then

     

    Solution

     

     

    In angular projection, the body at the highest point has velocity = u cosθ in the horizontal direction which makes zero angles with horizontal direction.

     

     

  • Question 5
    4 / -1

     

    Two particles are projected simultaneously in the same vertical plane, from the same point, both with different speeds and at different angles with horizontal. The path followed by one, as seen by the other, is

     

    Solution

     

     

    Letbe the initial velocities of the two particles and θ1​ and θ2​ be their angles of projection with the horizontal.
    The velocities of the two particles after time t are,



     

     

  • Question 6
    4 / -1

     

    A football is kicked into the air vertically upwards with velocity u. The velocity of the ball at the highest point is

     

    Solution

     

     

    The velocity of the ball at the highest point is zero.

     

     

  • Question 7
    4 / -1

     

     The equations of motion of a projectile are given by x = 36t m and 2y = 96t - 9.8t2 m. The angle of projection is

     

    Solution

     

     

    Given x = 36t
    and 2y = 96t − 9.8t2
    or y = 48t − 4.9t2
    Let the initial velocity of projectile be u and angle of projection θ. Then, Initial horizontal component of velocity,

     

     

  • Question 8
    4 / -1

     

    The relation between the time of flight of projectile Tf, and the time to reach the maximum height tm is

     

    Solution

     

     

    Time to reach maximum height = tm
    Time to reach back to ground = tm
    Total time of flight, Tf = tm + tm = 2tm

     

     

  • Question 9
    4 / -1

     

    A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with a speed of 15 m s-1. The time taken by the stone to reach the ground is

     

    Solution

     

     

    Here, u = 15ms−1

    h = 490m

    Time taken by the stone to reach the ground is

     

     

  • Question 10
    4 / -1

     

    The speed with which the stone hits the ground is

     

    Solution

     

     

    Motion along horizontal direction, ↓+ ve
    ux = 15ms−1, ax = 0 
    vx = ux + axt = 15 + 0 × 10 = 15ms−1
    Motion along vertical direction,
    uy = 0,ay = g
    vy = uy + ayt = 0 + 9.8 × 10 = 98ms−1
    ∴ Speed of the stone when it hits the ground is

     

     

  • Question 11
    4 / -1

     

    Two balls are projected at an angle θ and (90° - θ) to the horizontal with the same speed. The ratio of their maximum vertical heights is

     

    Solution

     

     

     

     

  • Question 12
    4 / -1

     

    Two particles are projected in air with speed u at angles θ1​ and θ2​ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then which one of the following is correct?
    (where T1​ and T2​ are the time of flight.)

     

    Solution

     

     

     

     

  • Question 13
    4 / -1

     

    The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 m s-1 without hitting the ceiling of the hall is

     

    Solution

     

     

    Here, u = 56 m s-1 Let θ be the angle of projection with the horizontal to have maximum range, with maximum height = 40 m

     

     

  • Question 14
    4 / -1

     

    The maximum horizontal distance covered by the ball will be

     

    Solution

     

     

     

     

  • Question 15
    4 / -1

     

    If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is:

     

    Solution

     

     

     

     

  • Question 16
    4 / -1

     

    When air resistance is taken into account while dealing with the motion of the projectile which of the following properties of the projectile, shows an increase?

     

    Solution

     

     

    In the presence of air resistance, the range, maximum height, speed at which the projectile strikes the ground will decreases whereas the angles at which the projectile strikes the ground will increases.

     

     

  • Question 17
    4 / -1

     

    Two projectiles are are fired from the same point with the same speed at angles 60° and 30° respectively. Which one of the follwing is true?

     

    Solution

     

     

    The horizontal range is the same when angle of projection is θ or (90∘-θ).

     

     

  • Question 18
    4 / -1

     

    Galileo writes that for angles of projection of a projectile at angle (45o + θ) and (45o − θ), the horizontal ranges described by the projectile are in the ratio of: (if θ ≤ 45o)

     

    Solution

     

     

     

     

  • Question 19
    4 / -1

     

     A cricket ball is thrown at a speed of 30ms−1 in a direction 30∘ above the horizontal. The time taken by the ball to return to the same level is
    (Take g = 10ms−2)

     

    Solution

     

     

    Here, u = 30ms−1,

    θ = 30∘,

    g = 10ms−2

    The time taken by the ball to return to the same level is

     

     

  • Question 20
    4 / -1

     

    Then the distance from the thrower to the point where the ball returns to the same level is

     

    Solution

     

     

    Given data,

    u = 30ms−1

    θ = 300

    g = 10ms−2

    The distance from the thrower to the point where the ball returns to the same level is

     

     

  • Question 21
    4 / -1

     

    The maximum height attained by the ball is

     

    Solution

     

     

    The maximum height is given by

     

     

  • Question 22
    4 / -1

     

    A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball?

     

    Solution

     

     

    Let u be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizontal, θ = 45∘. Then Rmax = u2/g = 100m
    If ball is projected vertically upwards (θ = 90∘) from ground then H attains maximum value.

    ∴  The height to which cricketer can through the ball is = Rmax/2 = 100/2 = 50m.

     

     

  • Question 23
    4 / -1

     

    An aeroplane flying horizontally with a speed of 360 km h-1 releases a bomb at a height of 490 m from the ground. If g = 9.8 m s-2, it will strike the ground at

     

    Solution

     

     

    Time taken by the bomb to fall through a height of 490 m

    Distance at which the bomb strikes the ground = horizontal velocity x time

     

     

  • Question 24
    4 / -1

     

    A ball is thrown from the top of a tower with an initial velocity of 10 m s-1 at an angle ot 30° with the horizontal. If it hits the ground at a distance of 17.3 m from the base of the tower, the height of the tower is (Take g = 10 m s-2)

     

    Solution

     

     

    The ball is thrown at an angle, θ = 30o.

    Initial velocity of the ball, u = 10 m/s 

    Horizontal range of the ball, R = 17.3 m
    We know that, R = u cosθ t,
     where t is the time of flight 

    using equation of motion we get:-

    ⟹ Height of tower, h = 10 m

     

     

  • Question 25
    4 / -1

     

    The speed of a projectile at its maximum height is √3/2 times its initial speed. If the range of the projectile is P times the maximum height attained by it, then P equals

     

    Solution

     

     

    Given √3/2 ​​u = u cos θ
    speed at maximum height or cos θ = √3/2 or θ = 30o

    Given that P Hmax ​=  R

    We know Hmax ​= R tan θ​/4

     

     

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