Chapter 9.5, Problem 37PPS

a.

To determine

To find: The time in which the object reaches the ground on each planet.

Answer to Problem 37PPS

Time to reach on mars = $8.021\sec$ .

Time to reach on earth = $4.938\sec$ .

Explanation of Solution

Given Information:It is given the height of an object after t second it dropped

  $\begin{array}{l}h=\frac{-1}{2}g{t}^2+h_o\\ where{h}_o=initialheight\\ g=accelerationduetogravity\\ g(mars)=3.73m/s^2\\ g(earth)=9.8m/s^2\\ h_o=120\end{array}$

Given the equation $h=\frac{-1}{2}g{t}^2+h_o$

For mars

  $h=0 ,G=3.73 ,h_0=120$ .

Put in equation

  $\begin{array}{l}0=\frac{-1}{2}(3.73)t^2+120\\ \frac{1}{2}(3.73)t^2=120\\ 3.73t^2=240\\ t^2=\frac{240}{3.73}\\ t^2=\frac{24000}{373}\\ =64.34\\ t=\pm \sqrt{64.34}\\ t=\pm 8.021\end{array}$

But time is never negative t = 8.021 seconds

For earth

  $h_0=120,H =0,G=9.8$

  $\begin{array}{l}0=\frac{-1}{2}(9.8)t^2+120\\ \frac{1}{2}(9.8)t^2=120\\ t^2=\frac{240}{9.8}\\ t^2=\frac{24000}{98}\\ =24.49\\ t=\pm \sqrt{24.49}\\ t=\pm 4.938\end{array}$

Thus, time to reach on mars $=8.021sec$ , time to reach on earth $=4.938sec$ .

b.

To determine

To find: The planet that would object reach the ground first.

Answer to Problem 37PPS

Object reach on earth first.

Explanation of Solution

Given Information: It is given the height of an object after t second it dropped

  $\begin{array}{l}h=\frac{-1}{2}g{t}^2+h_o\\ where{h}_o=initialheight\\ g=accelerationduetogravity\\ g(mars)=3.73m/s^2\\ g(earth)=9.8m/s^2\\ h_o=120\end{array}$

Given Information: It is given the height of an object after t second it dropped

  $\begin{array}{l}h=\frac{-1}{2}g{t}^2+h_o\\ where{h}_o=initialheight\\ g=accelerationduetogravity\\ g(mars)=3.73m/s^2\\ g(earth)=9.8m/s^2\\ h_o=120\end{array}$

Given the equation $h=\frac{-1}{2}g{t}^2+h_o$

For mars

  $h=0 ,G=3.73 ,h_0=120$ .

Put in equation

  $\begin{array}{l}0=\frac{-1}{2}(3.73)t^2+120\\ \frac{1}{2}(3.73)t^2=120\\ 3.73t^2=240\\ t^2=\frac{240}{3.73}\\ t^2=\frac{24000}{373}\\ =64.34\\ t=\pm \sqrt{64.34}\\ t=\pm 8.021\end{array}$

But time is never negative t = 8.021 seconds

For earth

  $h_0=120,H =0,G=9.8$

  $\begin{array}{l}0=\frac{-1}{2}(9.8)t^2+120\\ \frac{1}{2}(9.8)t^2=120\\ t^2=\frac{240}{9.8}\\ t^2=\frac{24000}{98}\\ =24.49\\ t=\pm \sqrt{24.49}\\ t=\pm 4.938\end{array}$

Thus, the object will reach the earth first.