Chapter 1, Problem 64A

To determine

To Find:The height of the ball from the graph at $t=0\text{s,}t=2\text{s and }t=5\text{s}$ .

Explanation of Solution

Given:

A height versus time graph showing the height of the ball for first $1.\text{5 s}$ .

Formula Used:

The height can be obtained by using the second equation of motion:

  $h=ut+\frac{1}{2}g{t}^2$

Where,h is the height

uis the initial velocity

Consider the acceleration due to gravity, $g=10{\text{m/s}}^{\text{2}}$

Calculation:

From the graph

At $t=0$ , $h=20\text{m}$

At $t=2\text{s}$

Velocity is zero at the maximum height when time is one second.

  $h=ut+\frac{1}{2}g{t}^2$

Considering top most point as the initial point for now, $u=0$

Height covered from the top at $t=1\text{s}$ till $t=2\text{s}$

  $\Delta t=\left(2-1\right)\text{s}=1\text{s}$

  $h=0+\frac{1}{2}10{(1)}^2$

  $h=5\text{m}$

Therefore, height at $t=2\text{s}$ is $h_2=2\text{5}\text{m}-5\text{m}=20\text{m}$

At $t=5\text{s}$

Velocity is zero at the top when time is one second.

  $h=ut+\frac{1}{2}g{t}^2$

Considering top most point as the initial point for now, $u=0$

Height covered from the top at $t=1\text{s}$ till $t=5\text{s}$ .

  $h=0+\frac{1}{2}10{(4)}^2$

  $h=80\text{m}$

But as the ball is descending from the maximum height of $25\text{m}$ the ball will be at the ground at $t=5\text{s}$ and therefore the height will be 0 m.

Conclusion:

At $t=0$ height, $h=20\text{m}$

At $t=2\text{s}$ height, $h=20\text{m}$

At $t=5\text{s}$ height, $h=0$