Chapter 9.4, Problem 43PPS

(a)

To determine

Find on which planet the object would reach the ground floor.

Answer to Problem 43PPS

The object on Earth will reach the ground first because it is falling at a faster rate.

Explanation of Solution

Given:

The height of an object t seconds after it is dropped is given by the equation $h=-\frac{1}{2}g{t}^2+h_0$ where $h_0$ the initial height is and g is the acceleration due to gravity. The acceleration due to gravity near the surface of Mars is $3.73\text{m}\text{/}{\text{s}}^{\text{2}}$ , while on Earth it is 9.8 $\text{m}\text{/}{\text{s}}^{\text{2}}$ . Suppose an object is dropped from an initial height of 120 meters above the surface of each planet.

Concept Used:

The acceleration due to gravity near the surface of Mars is $3.73\text{m}\text{/}{\text{s}}^{\text{2}}$ and the acceleration due to gravity near the surface of is 9.8 $\text{m}\text{/}{\text{s}}^{\text{2}}$ .

The object on Earth will reach the ground first because it is falling at a faster rate.

Thus, the object on Earth will reach the ground first because it is falling at a faster rate.

(b)

To determine

Find the time would it take the object to reach the ground on each planet.

Answer to Problem 43PPS

The time needed for Mars is $t\approx 8.0$ seconds and in Earth is $t\approx 4.9$ seconds

Explanation of Solution

Given:

The height of an object t seconds after it is dropped is given by the equation $h=-\frac{1}{2}g{t}^2+h_0$ where $h_0$ the initial height is and g is the acceleration due to gravity. The acceleration due to gravity near the surface of Mars is $3.73\text{m}\text{/}{\text{s}}^{\text{2}}$ , while on Earth it is 9.8 $\text{m}\text{/}{\text{s}}^{\text{2}}$ . Suppose an object is dropped from an initial height of 120 meters above the surface of each planet.

Concept Used:

Us the equation $h=-\frac{1}{2}g{t}^2+h_0$

The acceleration due to gravity near the surface of Mars is $3.73\text{m}\text{/}{\text{s}}^{\text{2}}$ ,

The acceleration due to gravity near the surface of is 9.8 $\text{m}\text{/}{\text{s}}^{\text{2}}$ .

Calculation:

Mars:

  $\begin{array}{l}0=-\frac{1}{2}(3.71)t^2+120\\ 0=-1.855t^2+120\\ -120=-1.855t^2+120-120[\text{subtract}\text{120}\text{from}\text{both}\text{sides}]\\ -120=-1.855t^2\\ \frac{-120}{-1.855}=\frac{-1.855t^2}{1.855}\\ t\approx \pm 8.0\end{array}$

So, $t\approx 8.0$ seconds.

Earth:

  $\begin{array}{l}0=-\frac{1}{2}·9.8t^2+120\\ 0=-4.9t^2+120\\ -120=-4.9t^2[\text{Subtract}\text{120}\text{from}\text{each}\text{side}]\\ \frac{-120}{-4.9}=\frac{-4.9t^2}{-4.9}\\ 24.49\approx t^2\\ t\approx \pm 4.9\end{array}$

So, $t\approx 4.9$ seconds.

Thus, the time needed for Mars is $t\approx 8.0$ seconds and in Earth is $t\approx 4.9$ seconds.

(c)

To determine

Explain the reason how long would it take the object to reach the ground seen reasonable.

Answer to Problem 43PPS

Yes, the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

Explanation of Solution

Given:

The height of an object t seconds after it is dropped is given by the equation $h=-\frac{1}{2}g{t}^2+h_0$ where $h_0$ the initial height is and g is the acceleration due to gravity. The acceleration due to gravity near the surface of Mars is $3.73\text{m}\text{/}{\text{s}}^{\text{2}}$ , while on Earth it is 9.8 $\text{m}\text{/}{\text{s}}^{\text{2}}$ . Suppose an object is dropped from an initial height of 120 meters above the surface of each planet.

Concept Used:

Yes, the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.

Thus, yes, the acceleration due to gravity is much greater on Earth than on Mars, so the time to reach the ground should be much less.