Solve the problem.
-The acceleration of gravity near the surface of Mars is $3.72 \mathrm {~m} / \mathrm { sec } ^ { 2 }$. If a rock is blasted straight up from the surface with an initial velocity of $85 \mathrm {~m} / \mathrm { sec }$ (about $190 \mathrm { mph }$ ), how high does it go? (Hint: When is velocity zero?)
A) Approximately 1942.25 meters
B) Approximately 22.85 meters
C) Approximately 971.1 meters
D) Approximately 170 meters

Question

Quizplus · Accepted Answer

We can use the kinematic equation:
$$ v^2 = u^2 + 2as $$
where
- $v$ is the final velocity (which is 0 m/s at the highest point)
- $u$ is the initial velocity (which is 85 m/s)
- $a$ is the acceleration due to gravity on Mars (which is 3.72 m/s$^2$)
- $s$ is the displacement (the height we are trying to find)

Plugging in the values, we get:

$$\begin{aligned} 0^2 &= (85 \mathrm{~m/s})^2 + 2 \cdot 3.72 \mathrm{~m/s^2} \cdot s \
0 &= 7225 - 7.44s \
7.44s &= 7225 \
s &= \frac{7225}{7.44} \approx \boxed{	extbf{(C) } 971.1 \mathrm{~m}} \end{aligned} $$

So the rock goes approximately 971.1 meters high.

Solve the Problem 3.72 m/sec23.72 \mathrm {~m} / \mathrm { sec } ^ { 2 }3.72 m/sec2

Solve the Problem $3.72 \mathrm {~m} / \mathrm { sec } ^ { 2 }$