Question

a ball is equipped with a speedometer and launched straight upward. the speedometer reading four seconds after launch is shown at the right; the ball is moving downward. at what approximate times would the ball be moving upward and display the following speedometer readings? time (s): tap to answer time (s): tap to answer

Explanation:

Step1: Identify the acceleration due to gravity

The acceleration due to gravity near the Earth's surface is approximately $g = 9.8\ m/s^{2}\approx10\ m/s^{2}$, and it acts down - ward. When the ball is moving upward, its velocity is decreasing at a rate of $g$ per second, and when moving downward, its velocity is increasing at a rate of $g$ per second.

Step2: Analyze the velocity at $t = 4\ s$

At $t = 4\ s$, the ball is moving downward and has a speed of $v_4= 10\ m/s$. Using the equation $v = v_0+at$, where $a=-g$ (negative when moving upward and positive when moving downward), and $v_0$ is the initial velocity. If we consider the motion from the maximum - height to $t = 4\ s$, and at the maximum - height $v = 0$. Using $v=v_{max}+at$, with $v_{max} = 0$ and $a = g$ and $t$ is the time from the maximum - height to $t = 4\ s$. So $10=0 + 10t_1$, then $t_1 = 1\ s$. The time to reach the maximum - height is $t_{max}=4 - 1=3\ s$.

Step3: Find the time for the first speedometer reading ($v = 10\ m/s$ moving upward)

Using $v=v_0 - gt$, when $v = 10\ m/s$ moving upward and $v_0$ is the initial velocity. We know that at $t = 3\ s$ the ball reaches the maximum - height ($v = 0$). Using $v=v_0 - gt$, and since $v_0=gt_{max}=30\ m/s$. When $v = 10\ m/s$ moving upward, $10 = 30-10t$, then $t = 2\ s$.

Step4: Find the time for the second speedometer reading ($v = 30\ m/s$ moving upward)

When $v = 30\ m/s$ moving upward, using $v=v_0 - gt$ with $v_0 = 30\ m/s$. Then $30=30 - 10t$, so $t = 0\ s$.

Answer:

The times are $0\ s$ and $2\ s$.