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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

15 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 10 s
Taking upward direction as $+ v e,$
$u = + 50 m / s a = - g m / s^{2} s = 0 [As the ball is reaching its initial point.]$
By second equation of motion,
$s = u t + \frac{1}{2} a t^{2} ⟹ 0 = 50 t + \frac{1}{2} (- 10) t^{2} ⟹ 0 = 50 - 5 t ⟹ t = 10 s e c .$

Suggest Corrections

Similar questions

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 the ball , how long does it take to return to his hands ?

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

Q. A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s¯¹. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s¯¹ and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ?

Q. A player throws a ball upwards with an initial speed of 30 m

s^{- 1}

. How long does the ball take to return to the player's hands? (Take g = 10 m

s^{- 2}

Q. A girl standing on a pedestal at rest throws a ball upwards with maximum possible speed of

50 m s^{- 1}

. If the platform starts moving up at

5 m s^{- 1}

and the girl again throws the similar ball in similar way. The time taken in the previous case and the time taken in the second case to return to her hands are

Join BYJU'S Learning Program

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).

The correct option is B 10 sTaking upward direction as +ve,u=+50m/sa=−g m/s2s=0[As the ball is reaching its initial point.]By second equation of motion,s=ut+12at2⟹0=50t+12(−10)t2⟹0=50−5t⟹t=10sec.

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 $s^{- 2}$ ).

The correct option is B 10 s
Taking upward direction as $+ v e,$
$u = + 50 m / s a = - g m / s^{2} s = 0 [As the ball is reaching its initial point.]$
By second equation of motion,
$s = u t + \frac{1}{2} a t^{2} ⟹ 0 = 50 t + \frac{1}{2} (- 10) t^{2} ⟹ 0 = 50 - 5 t ⟹ t = 10 s e c .$