Question

3 free fall (continued)

1. maximum height and flight time the free - fall acceleration on mars is about one - third that on earth. suppose you throw a ball upward with the same velocity on mars as on earth.

a. how would the balls maximum height compare to that on earth?
b. how would its flight time compare?

1. velocity and acceleration suppose you throw a ball straight up into the air. describe the changes in the velocity of the ball. describe the changes in the acceleration of the ball.
2. critical thinking a ball thrown vertically upward continues upward until it reaches a certain position, and then falls downward. the balls velocity is instantaneously zero at that highest point. is the ball accelerating at that point? devise an experiment to prove or disprove your answer.

Explanation:

Step1: Recall kinematic - equation for maximum height

The kinematic equation for maximum height $h$ of a vertically - thrown object is $v^{2}=v_{0}^{2}-2gh$, where at maximum height $v = 0$. So, $h=\frac{v_{0}^{2}}{2g}$. Let $g_{E}$ be the acceleration due to gravity on Earth and $g_{M}$ be the acceleration due to gravity on Mars. Given $g_{M}=\frac{1}{3}g_{E}$.

Step2: Calculate the ratio of maximum heights

Let $h_{E}=\frac{v_{0}^{2}}{2g_{E}}$ and $h_{M}=\frac{v_{0}^{2}}{2g_{M}}$. Then $\frac{h_{M}}{h_{E}}=\frac{g_{E}}{g_{M}}$. Substituting $g_{M}=\frac{1}{3}g_{E}$, we get $\frac{h_{M}}{h_{E}} = 3$. So the ball's maximum height on Mars is 3 times that on Earth.

Step3: Recall kinematic - equation for flight time

The kinematic equation for the time of flight $t$ of a vertically - thrown object is $v = v_{0}-gt$. For the upward - motion, at the maximum height $v = 0$, and the time to reach the maximum height $t_{up}=\frac{v_{0}}{g}$. The total flight time $t = \frac{2v_{0}}{g}$. Let $t_{E}=\frac{2v_{0}}{g_{E}}$ and $t_{M}=\frac{2v_{0}}{g_{M}}$.

Step4: Calculate the ratio of flight times

Then $\frac{t_{M}}{t_{E}}=\frac{g_{E}}{g_{M}}$. Substituting $g_{M}=\frac{1}{3}g_{E}$, we get $\frac{t_{M}}{t_{E}}=3$. So the ball's flight time on Mars is 3 times that on Earth.

Step5: Analyze velocity and acceleration of a thrown ball

When a ball is thrown straight up, its initial velocity $v_{0}$ is positive (upward). As it moves up, the velocity $v$ decreases due to the downward - acting acceleration $g$ (opposite to the direction of motion). At the maximum height, $v = 0$. As it falls back down, the velocity becomes negative (downward) and increases in magnitude. The acceleration of the ball is constant throughout the motion (neglecting air resistance) and is equal to the acceleration due to gravity $g\approx9.8\ m/s^{2}$ downward.

Step6: Analyze acceleration at the maximum height

The ball is accelerating at the highest point. The acceleration due to gravity is always acting on the ball near the surface of a planet. To prove this experimentally, one could use a high - speed camera to record the motion of the ball. By analyzing the frames of the video near the maximum height, one can observe the change in the ball's position over time. Using the kinematic equations and the data from the video (such as the position of the ball in consecutive frames), one can calculate the acceleration.

Answer:

a. The ball's maximum height on Mars is 3 times that on Earth.
b. The ball's flight time on Mars is 3 times that on Earth.
For question 50: The velocity of the ball decreases as it moves upward, becomes zero at the maximum height, and then increases in the downward direction. The acceleration of the ball is constant and equal to the acceleration due to gravity (downward) throughout the motion.
For question 51: The ball is accelerating at the highest point. An experiment could involve using a high - speed camera to record the ball's motion and then analyzing the data to calculate the acceleration near the maximum height.