A ball is thrown vertically upwards from the top of a tower with an initial velocity of - 19-6 nbsp-m nbsp-s-1- The ball reaches the ground after - 5 nbsp-s- Calculate - -i- the height of the tower- nbsp- -ii- the velocity of ball on reaching the ground- Take - g-9-8 nbsp-m nbsp-s-2- -i- 24-5 m- - -ii- 19-4- -160- m- - s-1- -i- 24-5 m- - -ii- 28- - m- - s-1- -i- 25 m- - -ii- 29-4- -160- m- - s-1- -i- 24-5 m- - -ii- 29-4- -160- m- - s-1-

A ball is thrown vertically upward from the top of a tower with an initial velocity of 19.6 metre per second. The ball reaches the ground after 5s. Calculate

A. The height of the tower

B. The velocity of ball reaching the ground

Q. A ball which is thrown vertically up from the top of the tower reaches the ground in

12

s. Another ball thrown vertically downwards from the same position with the same velocity takes

4

s to reach the ground. Find the height of the tower.

(T a k e g = 10 {m s}^{- 2})

Q. A ball is thrown vertically upwards. It returns

6 s

later. Calculate:

(i)

the greatest height reached by the ball, and

(i i)

the initial velocity of the ball. (Take

g = 10 m s^{- 2}

)

Join BYJU'S Learning Program

A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6ms−1. The ball reaches the ground after 5s. Calculate :(i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g=9.8ms−2.

The correct option is D (i)24.5m, (ii)29.4 m s−1s=ut+12at2 =19.6×5−12×9.8×52 =−24.5 mHence, height of tower =24.5mv=u+at =19.6−9.8×5=−29.4 m/s

A ball is thrown vertically upwards from the top of a tower with an initial velocity of $19.6 m s^{- 1}$ . The ball reaches the ground after $5 s$ . Calculate : $(i)$ the height of the tower, $(i i)$ the velocity of ball on reaching the ground. Take $g = 9.8 m s^{- 2}$ .

The correct option is D $(i) 24.5 m$ , $(i i) 29.4$ $m$ $s^{- 1}$
$s = u t + \frac{1}{2} a t^{2}$
$= 19.6 \times 5 - \frac{1}{2} \times 9.8 \times 5^{2}$
$= - 24.5 m$

Hence, height of tower $= 24.5 m$
$v = u + a t$
$= 19.6 - 9.8 \times 5 = - 29.4 m / s$