A girl standing on a stationary lift (open from above) throws a ball upwards with initial speed 50 m . The time taken by the ball to return to her hands is (Take g = 10 m .)

Question

A girl standing on a stationary lift (open from above) throws a ball upwards with initial speed 50 m $s^{- 1}$ . The time taken by the ball to return to her hands is (Take g = 10 m $s^{- 2}$ .)

Smart Achievers · Accepted Answer

(2)

$\begin{matrix} Using, S = u t + \frac{1}{2} a t^{2} \\ Here, S = 0, u = 50 m s^{- 1}, a = - g = - 10 m s^{- 2} \\ ∴ 0 = 50 t + \frac{1}{2} \times (- 10) t^{2} = 50 t - 5 t^{2} \Rightarrow t = 10 s \end{matrix} ↑ + v e$

Solution

Q.
A girl standing on a stationary lift (open from above) throws a ball upwards with initial speed 50 m $s^{- 1}$ . The time taken by the ball to return to her hands is (Take g = 10 m $s^{- 2}$ .)

1 5 s

2 10 s

3 15 s

4 20 s

Ans.
(2)

$\begin{matrix} Using, S = u t + \frac{1}{2} a t^{2} \\ Here, S = 0, u = 50 m s^{- 1}, a = - g = - 10 m s^{- 2} \\ ∴ 0 = 50 t + \frac{1}{2} \times (- 10) t^{2} = 50 t - 5 t^{2} \Rightarrow t = 10 s \end{matrix} ↑ + v e$

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