Question

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 v_x = 2.0 m/s, v_y = 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?

Answer 1

(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

\(\theta=\tan ^{-1}\left[\frac{4 \mathrm{~m} / \mathrm{s}}{2 \mathrm{~m} / \mathrm{s}}\right]\)

\(=\tan ^{-1}(2)\)

\(=63.43^{\circ}\)

Answer 2

Solving vy(1) and vy(2) by substitution shows
that g = 2 m/s^2 on Exidor.

vy(1) = 2 = vy - g * 1
vy(2) = 0 = vy - g * 2
g = 2 m/s

Find initial value of vy
v(1) = vy - gt
2 = vy - 2(1)
vy = 4 m/s
vx = 2 m/s = constant

g = 2 m/s^2
V(t) = 2i + (4-gt)j
V(0) = 2i + 4j m/s ->initial velocity
launch angle: Arctan(y,x) = 63.435 degrees

V(1) = 2i + 2j m/s
V(2) = 2i + 0j m/s
V(3) = 2i - 2j m/s