Question

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed = 49 m/s.
(A) How much time does the ball take to return to his hands?
(B) If the lift starts moving up with a uniform speed of 5 m/s and he again throws a ball, how long does it take to return to his hand?

Answer 1

Hint: When a boy throws a ball in lift and that ball returns in his hand, the displacement will be zero because the initial position and final position of the ball is the same. And acceleration will be working in the opposite direction of velocity, hence acceleration will be negative.

Complete Step by step solution
A: When a lift is stay, taken time by ball to return in his hand
Given data: Displacement, S = 0
Initial Velocity, u = 49 m/s
Acceleration = - g
Taken time, t =?
Hence, according the 3rd equation of linear motion
 $\Rightarrow S=ut+{\displaystyle \frac{1}{2}}a{t}^2$
 $\Rightarrow 0=(49)t+{\displaystyle \frac{1}{2}}(-g)t^2$ …………………….. Put the values in that formula
 $\Rightarrow 0=49t-{\displaystyle \frac{1}{2}}g{t}^2$
 $\Rightarrow 49t={\displaystyle \frac{1}{2}}g{t}^2$
 $\Rightarrow {\displaystyle \frac{49\times 2}{g}}={\displaystyle \frac{t^2}{t}}$
 $\Rightarrow {\displaystyle \frac{49\times 2}{9.8}}=t$
 $\Rightarrow t=10\sec$
Hence, the ball will take time 10 second to return in his hand.
B: When a lift starts move, taken time by ball to return in his hand
Given data:
Lift moving with uniform speed, v = 5 m/s
Initial Velocity, u = 49 m/s
Relative initial velocity, $u=(49+5)t^{\prime }m/s$
 $\Rightarrow u=54t^{\prime }m/s$
Displacement, $S=5t^{\prime }$
Acceleration = - g
Taken time, t =?
Hence, according the 3rd equation of linear motion
 $\Rightarrow S=ut+{\displaystyle \frac{1}{2}}a{t}^2$
 $\Rightarrow 5t^{\prime }=54t^{\prime }+{\displaystyle \frac{1}{2}}(-g)t{}^{\prime 2}$ …………………….. Put the values in that formula
 $\Rightarrow 54t^{\prime }-5t^{\prime }={\displaystyle \frac{1}{2}}g{t}^2$
 $\Rightarrow 49t^{\prime }={\displaystyle \frac{1}{2}}gt{}^{\prime 2}$
 $\Rightarrow 49={\displaystyle \frac{g}{2}}{t}^{\prime }$
 $\Rightarrow {\displaystyle \frac{49\times 2}{g}}=t^{\prime }$
 $\Rightarrow {\displaystyle \frac{49\times 2}{9.8}}=t^{\prime }$
 $\Rightarrow t^{\prime }=10\sec$
Hence, the ball will take time 10 seconds to return in his hand during lift moves.

Note
If the body moves with uniform velocity, they will not have acceleration, hence, this type phenomenon is called ‘inertial frame of reference’ otherwise we can say, the motion of particles not subjected to influence force that particles move in a straight line at a constant speed. A frame of reference in which Newton’s laws of motion hold well is called an inertial frame of reference. Such a frame is an uncelebrated frame of reference.