Question

1. pinkie pie shoots her party cannon from a height of 0.5m. if it has a range of 4m, a. what is the final vertical velocity of the party supplies? b. what is the initial horizontal velocity?

Explanation:

Step1: Analyze vertical - motion for part a

Use the kinematic equation $v_y^2 = v_{0y}^2+ 2a_y\Delta y$. Assume the party - supplies are in free - fall, $a_y=-g=- 9.8m/s^2$, the initial height $y_0 = 0.5m$ and the final height $y = 0m$. Assuming the initial vertical velocity $v_{0y}=0m/s$.
$v_y^2=0 + 2\times9.8\times0.5$

Step2: Calculate the final vertical velocity for part a

$v_y=\sqrt{2\times9.8\times0.5}=\sqrt{9.8}\approx3.13m/s$ (downward, so the vector has a negative sign in a coordinate system where up is positive).

Step3: Analyze horizontal - motion for part b

In the horizontal direction (assuming no air - resistance), $a_x = 0m/s^2$. The range $R = 4m$. First, find the time of flight $t$ from the vertical motion. Using the equation $y - y_0=v_{0y}t+\frac{1}{2}a_yt^2$. Since $v_{0y} = 0m/s$, $y - y_0=- 0.5m$ and $a_y=-9.8m/s^2$.
$-0.5=0\times t-\frac{1}{2}\times9.8\times t^2$
$4.9t^2 = 0.5$
$t=\sqrt{\frac{0.5}{4.9}}\approx0.32s$

Step4: Calculate the initial horizontal velocity for part b

In the horizontal direction, $x = v_{0x}t$. Given $x = R = 4m$ and $t\approx0.32s$.
$v_{0x}=\frac{4}{0.32}=12.5m/s$

Answer:

a. $v_y\approx - 3.13m/s$
b. $v_{0x}=12.5m/s$