Chapter 11.2, Problem 30SSC

To determine

To Compare: The potential energy of ball at half of total time of fall with half of total potential energy.

Answer to Problem 30SSC

The potential energy is ball at half of total time of fall is greater than the exactly half of total potential energy.

Explanation of Solution

Given:

Height, $h$ =20 m

Initial speed, $u$ = 0 m/s

Formula used:

Potential energy, $P=mgh$

Here, m is the mass, g is the acceleration due to gravity and h is the height.

Calculation:

Let, the mass of ball is $m$ .

The half of total potential energy is calculated as

  $\begin{array}{l}{P}_{1/2}=\frac{mgh}{2}\\ P_{1/2}=\frac{mg\left(20\right)}{2}\\ P_{1/2}=10mg\text{ (1)}\end{array}$

The total time of fall can be calculated using second equation of motion:

  $H=uT+\frac{g{T}^2}{2}$

Here, H is the displacement, u is the initial velocity, T is the time and gis the acceleration due to gravity.

  $\begin{array}{l}H=uT+\frac{g{T}^2}{2}\\ 20=\left(0\right)T+\frac{\left(9.8\right)T^2}{2}\\ T=\sqrt{\frac{20\times 2}{9.8}}\\ T=2.02\text{ s}\end{array}$

The half of total time of fall is

  $\begin{array}{l}{T}_{1/2}=\frac{2.02}{2}\\ T_{1/2}=1.01\text{ s}\end{array}$

The distance covered in $1.01$ s is

  $\begin{array}{l}{T}_{1/2}=\frac{2.02}{2}\\ T_{1/2}=1.01\text{ s}\\ h=ut+\frac{g{t}^2}{2}\\ h=\left(0\right)\left(1.01\right)+\frac{9.8{\left(1.01\right)}^2}{2}\\ h=4.99\text{ m}\end{array}$

The height of ball at $1.01$ s from the ground is

  $\begin{array}{l}{h}_g=20-4.99\\ h_g=15.01\text{ m}\end{array}$

Now, the potential energy of ball at $1.01$ s is as

  $\begin{array}{l}{P}_{1.01\text{ s}}=mg{h}_g\\ P_{1.01\text{ s}}=15.01mg\mathrm{...}\text{(2)}\end{array}$

On comparing equation (1) and (2),

  $P_{1.01\text{ s}}>P_{1/2}$

Conclusion:

The potential at half of time of fall is more than the exactly half the total potential energy of the ball.