Question

A ball is dropped from the top of A tower and reaches the ground in 4 second. The height of the tower is( g= 10m/s^2)

Answer 1

Assume the ball is dropped from rest. Then,

• initial velocity, $\displaystyle\sf{u=0}$

• acceleration, $\displaystyle\sf{a=g=10\ m\ s^{-2}}$

• time taken, $\displaystyle\sf{t=4\ s}$

Here the height from which the ball is dropped is the displacement, $\displaystyle\sf{s.}$

Then, by second kinematic equation,

$\displaystyle\longrightarrow\sf{s=ut+\dfrac{1}{2}at^2}$

$\displaystyle\longrightarrow\sf{s=0\times4+\dfrac{1}{2}\times10\times4^2}$

$\displaystyle\longrightarrow\sf{\underline{\underline{s=80\ m}}}$

Answer 2

Answer: 39.2metre

Explanation:

speed = distance/time    therefore [speed = acceleration due to gravity ]

9.8m/s²=distance/4second                

9.8 x 4=distance

39.2m = distance