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)

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) (A) 5 s (B) 10 s (C) 15 s (D) 20 s

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/53a23fca49fbb31772f6c1bb8dd06e7b-.jpeg" /> Using, S=ut+(1/2) at2 Here S = 0, u = 50 m s-1, a=-g=-10 m s-2 ∴ 0=50t+(1/2)×(-10)t2=50t-5t2 ⇒ t=10 s

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 $g = 10 m s^{- 2}$ )

$5 s$

$10 s$

$15 s$

$20 s$

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

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

5s

10s

15s

20s