Question

12. A physics student on Planet Exidor throws a ball, and it follows the parabolic trajectory shown in FIGURE EX4.12. The ball's position is shown at 1s intervals until$t=3\text{s}$. At $t=1\text{s}$, the ball's velocity is localid="1647396327258" $\overrightarrow{v}=(2.0\hat{i}+2.0\hat{j})\text{m}/\text{s}$.

a. Determine the ball's velocity at $t=0\text{s},2\text{s}$, and $3\text{s}$.

b. What is the value of g on Planet Exidor?

c. What was the ball's launch angle?

Short answer

Thus, the vertical component of the velocity of the ball at is$-2.0\text{m}/\text{s}$

Step-by-step solution

step 1. Given information

The ball's velocity at $t=1\text{s}$is data-custom-editor="chemistry" $\overrightarrow{v}(1\text{s})=(2.0\widehat{i}+2.0\widehat{j})\text{m}/\text{s}$.

The following figure shows the motion of the ball at each 1s interval.

Step 2. Explanation

Formula to calculate the vertical component of the velocity of the ball at $t=2\text{s}$is

$v_{y(2\text{s})}=v_{y(1\text{s})}+(-g)\left(t_{(2\text{s})}-t_{(1\text{s})}\right)$

Here, $v_{y(2\text{s})}$is the vertical component of the velocity of the ball at $t=2\text{s}$.

At $t=2\text{s}$, the velocity is${v_y}_{2s}=0\text{m}/\text{s}$

role="math" localid="1647477509413" $\begin{array}{l}0\text{m}/\text{s}=2.0\text{m}/\text{s}+(-g)(2\text{s}-1\text{s})\\ g(1\text{s})=(2.0\text{m}/\text{s)}\\ g\text{\hspace{0.17em}=\hspace{0.17em}2}.0\text{m}/\text{s}\end{array}$

Thus, the acceleration due to the gravity of the planet Exidor is $2\text{m}/{\text{s}}^2$.

The vertical component of the velocity of the ball at $t=1\text{s}$is

$\begin{array}{l}{v}_{y(1\text{s})}=v_{y(0\text{s})}+(-g)\left(t_{(1\text{s})}-t_{(0\text{s})}\right)\\ 2.0\text{m}/\text{s}=v_{y(0\text{s})}+\left(-2\text{m}/{\text{s}}^2\right)(1\text{s})\\ v_{y(0\text{s})}=2.0\text{m}/\text{s}+2.0\text{m}/\text{s}=4.0\text{m}/\text{s}\end{array}$

Thus, the vertical component of the initial velocity of the ball is $4.0\text{m}/\text{s}$.

Step 3. Explanation

The vertical component of the velocity of the ball t = 1 s at is

$\begin{array}{l}{v}_{y(1\text{s})}=v_{y(0\text{s})}+(-g)\left(t_{(1\text{s})}-t_{(0\text{s})}\right)\\ 2.0\text{m}/\text{s}=v_{y(0\text{s})}+\left(-2\text{m}/{\text{s}}^2\right)(1\text{s})\\ v_{y(0\text{s})}=2.0\text{m}/\text{s}+2.0\text{m}/\text{s}=4.0\text{m}/\text{s}\end{array}$

Thus, the vertical component of the initial velocity of the ball is $4.0\text{m}/\text{s}$.

The vertical component of the velocity of the ball at t = 2s is

$\begin{array}{l}{v}_{y(2\text{s})}=v_{y(0\text{s})}+(-g)\left(t_{(2\text{s})}-t_{(0\text{s})}\right)\\ 2.0\text{m}/\text{s}=v_{y(0\text{s})}+\left(-2\text{m}/{\text{s}}^2\right)(2\text{s})\\ v_{y(0\text{s})}=4.0\text{m}/\text{s}-4.0\text{m}/\text{s}=0.0\text{m}/\text{s}\end{array}$

Thus, the vertical component of the initial velocity of the ball is $0.0\text{m}/\text{s}$.

The vertical component of the velocity of the ball at t = 3s is

$\begin{array}{l}{v}_{y(3\text{s})}=v_{y(0\text{s})}-g\left(t_{(3\text{s})}-t_{(0\text{s})}\right)\\ v_{y(3\text{s})}=4\text{m}/\text{s}-\left(2\text{m}/{\text{s}}^2\right)\left(3\text{s}\right)\\ v_{y(3\text{s})}=-2.0\text{m}/\text{s}\end{array}$

Thus, the vertical component of the velocity of the ball at islocalid="1650257973632" $-2.0\text{m}/\text{s}$