Question

A body at rest is dropped from the top of a tower. Draw a displacement-time graph of its free fall under gravity during the first 06 seconds. Show table for displacement and time starting from t = 0 with an interval of 01 second. Take $g=10m{s}^{-2}.$

Answer 1

The body is at rest before it is being dropped from the tower. So the displacement of the body at any point is given by the equation

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

where,
u = initial velocity, g = acceleration due to gravity and t = time taken.

Initial velocity $u=0,$ Acceleration $a=10m/s^2$
Therefore, when the body is at rest, t = 0, then
$s_0=0$

The displacement covered is calculated for every increase of 1 second.
At $t=1s$, $s_1=0+\frac{1}{2}\left(10\right)(1{)}^2=5m$
At $t=2s$, $s_2=0+\frac{1}{2}\left(10\right)(2{)}^2=20m$
At $t=3s$, $s_3=0+\frac{1}{2}\left(10\right)(3{)}^2=45m$
At $t=4s$, $s_4=0+\frac{1}{2}\left(10\right)(4{)}^2=80m$
At $t=5s$, $s_5=0+\frac{1}{2}\left(10\right)(5{)}^2=125m$
At $t=6s$, $s_6=0+\frac{1}{2}\left(10\right)(6{)}^2=180m$

Therefore, the points corresponding to displacement $s$ every second $t$ are as follows:

$t=0,s=0m$

$t=1,s=5m$

$t=2,s=20m$

$t=3,s=45m$

$t=4,s=80m$

$t=5,s=125m$

$t=6,s=180m$

These points have been plotted as a displacement-time graph of free fall under gravity during the first 06 seconds.