Motion in A Straight Line Test - 12

Suppose velocity of projection = u

Then velocity at height h will be \(\mathrm{v=\sqrt{u^2-2gh}}\)

Putting \(\mathrm{h=10m,g=10m/s^2}\)

We get, \(\mathrm{\sqrt{{u^2}-200}}\)  .... (1)

but at t = 1 second is the time taken to reduce u to v

so, v − u − g(1) = u − g

u = v + g

Putting this in equation 1

\(\mathrm{v^ 2 =u^ 2-200=(v+g)^ 2-200}\)

\(\Rightarrow \) \(\mathrm{v^ 2 =v^ 2 +g ^2 +2vg-200}\)

\(\Rightarrow \) 0 = 100 + 2v(10) − 200

\(\Rightarrow \) 0 = −100 + 20v

v = 5m/s

\(v^2 = u^2 + 2as\)

Substituting u = 15 m/s, v = 0 m/s,

A = g = -10 \(m/s^2\)

\(15^2 = 2 \times 10 \times s\)

s = \(\frac{225}{20} = 11.25 m\)

Here we should use Newton’s second eqn. of motion 

That is

\(v^2 - u^2\) = 2gh

So here u = 19.6m/s and v = 0 m/s as after reaching a maximum height v becomes 0.

Now the eqn. changes to \(-u^2\) = 2gh which can be written as h = \(\frac{-u^2}{2g}\)

And here even g is also negative as the ball is thrown upwards

So, h = \(\frac{-384.16}{2 \times -9.8}\)

h = 19.6m

If \(t_1\) and \(t_2\) are the time when the body is at the same height then h

h = \(\frac{1}{2} \times g \times t_1 \times t_2\) 

h = \(\frac{1}{2} \times g \times 2 \times 10\)

h= 10 g

(i) Let the initial velocity of the body = u

(ii) And final velocity at the initial point is v.

(iii) Now the potential energy of the body at the point of projection is the same in both the upward and the downward journey.

(iv) Since the total energy of the system has to be conserved, the kinetic energies at that point must also be the same in both the journeys.

Free fall ⇒ falling with constant acceleration = g

The initial velocity with respect to ground will be 10 + 5 = 15 m/s

After 2sec its velocity will be v = 15 − gt =15 − 10 × 2 = −5 m/s

Whose value is 5 m/s

Note: Once the stone is projected from balloon it will not have any impact of the balloon after that moment.

Initial speed of the stone u = 0m/s

Time taken by the stone to reach the ground t = 4s

Height of the tower

H = \(ut + \frac{1}{2}at^2\)

\(\therefore H = 0 + \frac{1}{2} \times 10 \times 4^2\)

⇒ H = 80 m

u = 100 m/s

⇒ v = 0 m/s

⇒ g = - 10 \(m/s^2\)

We should find the approximate time taken to reach the ground

So, we know that,

v = u - gt

Implies,

t = \(\frac{v - u}{ -g}\)

t = \(\frac{0 - 100 }{ - 10}\)

t = 10 seconds.

Therefore the approximate time taken to reach the ground is 10 s + 10s = 20s seconds.

\(\because \vec v_{S.B} = \vec v_S - \vec v_B\)

\(\because \vec v_S = \vec v_{S.B} + \vec v_B\)

= \(10 \hat j + 5 \hat j = 15 \hat j\)

Velocity of stone w.r.t. ground = 15 m/s upward

v = u + at

⇒ v = 15 - 10 x 2

⇒ v = 15 - 20

⇒ v = - 5 m/s

Path of a particle moving under the influence of a force fixed in magnitude and direction towards a point is straight line because direction never change.

Initial speed u = 0; final speed v = 144 km/hr = \(144 \times \frac{5}{18}\) = 40 m/s;

Acceleration a is obtained from the formula, 

"v = u + at", i.e. 

a = \(\frac{v}{t}\)

= \(\frac{40}{20}\)

= 2 \(m/s^2\)

Distance covered in 20 s:

S = ut + \(\frac{1}{2}\) \(at^2\)

= \(\frac{1}{2} \times 2\times 20\times20\)

= 400 m

If \( t_2\) and \(2t_2\) are the time taken by particle to cover first and second half distance, respectively, then

\(t_1 = \frac{\frac{x}{2}}{3} =\frac{x}{6} \) 

\(x_1 = 4.5 t_2\) and \(x_2 = 7.5 t_2\) 

So, \(x_1 = x_2 = \frac{x}{2}\)

⇒ \(4.5 t_2 + 7.5t_2 = \frac{x}{2}\)

\(t_2 = -\frac{x}{24}\)

Total time \(t_1 + 2t_2 = \frac{x}{6} + \frac{x}{12} = \frac{x}{4}\)

So, average speed 4 m/sec.

Particle going A to M, distance traveled = \(\frac{s}{2}\)

\(v^2 = 2^2 + \frac{2as}{2}\)

⇒ \(v^2 - 4 = as \)

\(14^2 - v^2 = \frac{2as}{2}\)

⇒ \(196 - v^2 = as\)  

\(v^2 - 4 = 196 - v^2\)

⇒ v = 10 m/s Now,

\(t_1 = \frac{v - 2}{a} = \frac{10 - 2}{a} = \frac{8}{a}\) (Time from A to M)

\(t_2 = \frac{14 - v}{a} = \frac{14 - 10}{a} = \frac{4}{a}\)  (Time from M to B)

⇒ \(\frac{t_2}{t_1} = \frac{4/a}{8/a} = \frac{1}{2}\)

\(S_n = u + \frac{a}{2}(2n - 1) \) or, \(S = \frac{a}{2}(2 \times 2 - 1)\)

⇒ \(a = \frac{2}{3}sm/s^2\)

\(v = \sqrt{3x + 16}\) ⇒ \(v^2 = 3x + 16 \)

⇒ \(v^2 - 16 = 3x\) 

Comparing with \(v^2 - u^2 = 2aS\), we get,

u = 4 units, 2a = 3 or a = 1.5 units

Distance covered by a particle is zero only when it is at rest. Therefore, its displacement must be zero.

Height attained by balls in 2 sec is = \(\frac{1}{2} \times 9.8 \times 4 = 19.6 m\)

The same distance will be covered in 2 second (for descent) Time interval of throwing balls, remaining same.

So, for two balls remaining in air, the time of ascent or descent must be greater than 2 seconds.

Hence speed of balls must be greater than 19.6 m/sec.