A player throws a ball upwards with an initial speed of 29.4 ms -1. Taking g=9.8 ms -2, the time taken by the ball to return to the player's hand is

Question

A player throws a ball upwards with an initial speed of 29.4 ms -1. Taking g=9.8 ms -2, the time taken by the ball to return to the player's hand is (A) 3 s (B) 6 s (C) 9 s (D) 12 s

Tardigrade Admin · Accepted Answer

Using, s=u t+(1/2) a t2 s=0, u=29.4 ms -1 and a=-9.8 ms -2 0=29.4 t-(9.8/2) t2 0=t(29.4-4.9 t) ⇒ t = 0 s or t=(29.4/4.9)=6 s 0 s is the initial time, so the ball takes 6 s to return to the player's hand.

Q. A player throws a ball upwards with an initial speed of $29.4 m s^{- 1}$ . Taking $g = 9.8 m s^{- 2}$ , the time taken by the ball to return to the player's hand is

3 s

6 s

9 s

12 s

Using, $s = u t + \frac{1}{2} a t^{2}$
$s = 0, u = 29.4 m s^{- 1}$ and $a = - 9.8 m s^{- 2}$
$0 = 29.4 t - \frac{9.8}{2} t^{2}$
$0 = t (29.4 - 4.9 t) \Rightarrow t = 0 s$
or $t = \frac{29.4}{4.9} = 6 s$
$0 s$ is the initial time, so the ball takes $6 s$ to return to the player's hand.

Q. A player throws a ball upwards with an initial speed of 29.4ms−1. Taking g=9.8ms−2, the time taken by the ball to return to the player's hand is

3 s

6 s

9 s

12 s

Using, s=ut+21​at2 s=0,u=29.4ms−1 and a=−9.8ms−2 0=29.4t−29.8​t2 0=t(29.4−4.9t)⇒t=0s or t=4.929.4​=6s 0s is the initial time, so the ball takes 6s to return to the player's hand.

Q. A player throws a ball upwards with an initial speed of $29.4 m s^{- 1}$ . Taking $g = 9.8 m s^{- 2}$ , the time taken by the ball to return to the player's hand is

Using, $s = u t + \frac{1}{2} a t^{2}$
$s = 0, u = 29.4 m s^{- 1}$ and $a = - 9.8 m s^{- 2}$
$0 = 29.4 t - \frac{9.8}{2} t^{2}$
$0 = t (29.4 - 4.9 t) \Rightarrow t = 0 s$
or $t = \frac{29.4}{4.9} = 6 s$
$0 s$ is the initial time, so the ball takes $6 s$ to return to the player's hand.