Question

A body is dropped from a certain height at time $T=0$. It undergoes freefall and reaches the ground at time $T=t$ s. What is the expression for the height $h$ from which the body started its free fall?

Acceleration due to gravity is $g$.

• A. $h=g{t}^2$
• B. $h=gt$
• C. $h=\frac{1}{2}g{t}^2$
• D. $h=\left(\frac{1}{2}\right)g{t}^{-2}$

Answer 1

Answer: (C)

$h=\frac{1}{2}g{t}^2$

The distance covered with initial velocity $u$ and undergiong constant accleration $a$ in time $t$ is given by:

$s=ut+\frac{1}{2}a{t}^2$

In this case the distance travelled is $h$, acceleration is $g$

We know that $h=ut+\left(\frac{1}{2}\right)g{t}^2$.

Since the body is initially at rest, $u=0$.

Therefore, $h=\left(\frac{1}{2}\right)g{t}^2$