Chapter 3.3, Problem 50SSC

(a)

To determine

To Compare: The maximum height of the ball achieved on Mars with that on the Earth.

Explanation of Solution

Given:

Ball is thrown on Mars and Earth with the same velocity.

The free-fall acceleration on Mars is one-third of that on the Earth.

Formula Used:

Third equation of motion:

  $v^2-u^2=2as$

Where, v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.

Calculations:

The final velocity at the maximum height is zero, $v=0$ .

The initial velocity is same. $u_E=u{M}_{}=u$

Acceleration due to gravity on Mars $g_M=\frac{1}{3}{g}_E$

Where, $g_E$ is the acceleration due to gravity on Earth.

Let the height achieved on Mars be $h_M$ and that on Earth be $h_E$ .

For Mars:

  $\begin{array}{l}0-u_M{}^2=-2g_M{h}_M\\ \Rightarrow h_M=\frac{u^2}{2g_M}\end{array}$

For Earth:

  $\begin{array}{l}0-u_E{}^2=-2g_E{h}_E\\ \Rightarrow h_E=\frac{u^2}{2g_E}\end{array}$

Take the ratio:

  $\begin{array}{l}\frac{h_M}{h_E}=\frac{g_E}{g_M}\\ \Rightarrow h_M=3h_E\end{array}$

Conclusion:

Thus, the height achieved by ball on Mars would be three times to that on the Earth.

(b)

To determine

To Compare: The flight time of the ball on Mars with that on the Earth.

Explanation of Solution

Given:

Ball is thrown on Mars and Earth with the same velocity.

The free-fall acceleration on Mars is one-third of that on the Earth.

From previous part, the height achieved by ball on Mars would be three times to that on the Earth.

Formula Used:

The relation between the time of flight and maximum height achieved is:

  $T=2\sqrt{\frac{2h}{g}}$

Where, T is the time of flight, h is the maximum height achieved and g is the acceleration due to gravity.

Calculations:

Time of flight of ball on Mars:

  $T_M=2\sqrt{\frac{2h_M}{g_M}}$

Time of flight of ball on Earth:

  $T_E=2\sqrt{\frac{2h_E}{g_E}}$

Take the ratio:

  $\begin{array}{l}\frac{T_M}{T_E}=\frac{2\sqrt{\frac{2h_M}{g_M}}}{2\sqrt{\frac{2h_E}{g_E}}}\\ \Rightarrow \frac{T_M}{T_E}=\sqrt{\frac{h_M{g}_E}{h_E{g}_M}}\\ \because h_M=3h_E\text{and}{g}_M=\frac{1}{3}{g}_E\\ \Rightarrow \frac{T_M}{T_E}=\sqrt{\frac{3h_E{g}_E}{h_E\left(\frac{1}{3}\right)g_E}}\\ \Rightarrow \frac{T_M}{T_E}=\sqrt{9}\\ \Rightarrow T_M=3T_E\end{array}$

Conclusion:

Thus, the time of flightof ball on Mars would be three times to that on the Earth.