A physics student on Planet Exidor throws a ball, and it follows the parabolic trajectory shown in the figure. The ball's position is shown at 1.0 s intervalsuntil t = 3.0 s. At t = 1.0 s, the ball's velocity has components vx = 2.0 m/s, vy = 2.0 m/s.
a. Determine the x- and y- components of the ball's velocity at t = 0.0 s, 2.0 s, and 3.0 s.
b. What is the value of g on Planet Exidor?
c. What was the ball's launch angle?
(a) Throughout the motion, the horizontal component of velocity remains constant. Thu:
\(\bar{v}=v_{x} \hat{i}+v_{j} \hat{j} v_{x}\) is constant
At \(t=1\) ss \(s\) st \(t\) \(\bar{v}=(2 \hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}\)
Thus \(v_{z}=2\) and \(v_{y}=2\) at \(t=1 \mathrm{~s}\)
When the ball reaches point \(2,\) it is at the maximum height However, at the maximum
At \(t=28\) ss \(s=2\) \(\bar{v}=2 \hat{i}+0 \hat{j}\)
$$ \begin{aligned} &=2 \hat{i} \\ v_{j} &=0 \text { at } t=2 . \end{aligned} $$
Let the \(y\) -component of velocity at \(t=0\) be \(v_{v}\), then we have \(v_{y}=v_{v y}+g t\)
Whent \(=1 \mathrm{~s}, v_{y}=2 \mathrm{~m} / \mathrm{s}\)
\(2 \mathrm{~m} / \mathrm{s}=v_{0 y}+g(1 \mathrm{~s})\)
\(v_{v_{y}}=(2 \mathrm{~m} / \mathrm{s})-g(1 \mathrm{~s}\)
\(41 \mathrm{so},\) when \(t=2 \mathrm{~s}, v_{y}=0\)
$$ 0=v_{0 y}+g(2 s) $$
\(v_{0 y}=-g(2 s)\)
\(g=\frac{-v_{0 y} y}{2 \mathrm{~s}}\)
Thus,
\(\frac{v_{0 y}}{2}=2 \mathrm{~m} / \mathrm{s}\)
\(v_{b y}=4 \mathrm{~m} / \mathrm{s}\)
\(t=0\)
$$ \bar{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{i}+(4 \mathrm{~m} / \mathrm{s}) \hat{j} $$
At \(t=3 \mathrm{~s} \mathrm{~s}=\mathrm{s}==3 \mathrm{~s}=\)
\(v_{y}=v_{0 y}+g(3 \mathrm{~s})\)
$$ \begin{array}{l} =v_{0 y}+\left(\frac{-v_{y_{y}}}{2 \mathrm{~s}}\right)(3 \mathrm{~s}) \\ =v_{0 y}-\frac{3 v_{\mathrm{og}}}{2} \\ =-\frac{v_{0 y}}{2} \end{array} $$
$$ =-\frac{(4 \mathrm{~m} / \mathrm{s}}{2} $$
\(t=3 s \quad \bar{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{i}+(-2 \mathrm{~m} / \mathrm{s}) \hat{j}\)
at \(t=0 \mathrm{~s} \quad \bar{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{i}+(4 \mathrm{~m} / \mathrm{s}) \hat{j}\)
\(\begin{array}{ll}\mathrm{at} t=2 \mathrm{~s} & \bar{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{i}\end{array}\)
\(\begin{array}{ll}\text { at } t=3 \mathrm{~s} & \bar{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{i}-(2 \mathrm{~m} / \mathrm{s}) \hat{j}\end{array} \hat{j}\)
(b) We have
\(=-\frac{(4 m / s)}{(2 s)}\)
$$ =-2 \mathrm{~m} / \mathrm{s}^{2} $$
The value of \(g\) on planet Exidor is \(g=2 \mathrm{~m} / \mathrm{s}\) downward
\((c)\) We have
\(\theta=\tan ^{-1}\left[\frac{4 \mathrm{~m} / \mathrm{s}}{2 \mathrm{~m} / \mathrm{s}}\right]\)
\(=\tan ^{-1}(2)\)
\(=63.43^{\circ}\)
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