(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An arrow is shot with an initial velocity, V0, at an angle, [tex]\Theta[/tex], with respect to the horizontal. It reaches a maximum height of 2m above the launch point before hitting its target. The target is 15m horizontally, and 1m below the launch point.

Find the time in the air.

2. Relevant equations

y - y0 = Vy0 * t + 1/2 * ay * t^2

Vy = Vy0 + ay * t

x - x0 = Vx0 * t

3. The attempt at a solution

I started with some tables over the interval from launch till the arrow reaches its maximum:

y0 = 0m

y = y0 + 2m

Vy0 = V0 * sin[tex]\Theta[/tex] (m/s)

Vy = 0 (m/s)

ay = -9.81 (m/s^2)

t = unknown.

x0 = 0m

x = unknown.

Vx0 = V0 * cos[tex]\Theta[/tex]

t = unknown.

no acceleration

Started with short Y velocity equation:

0 = V0 * sin[tex]\Theta[/tex] - 9.81 * t

t = (v0 * sin[tex]\Theta[/tex]) / 9.81

Substituted into the long Y position equation:

2 = V0 * sin[tex]\Theta[/tex] * t - 4.905 * t^2

2 = V0 * sin[tex]\Theta[/tex] * ( (V0 * sin[tex]\Theta[/tex]) / 9.81 ) - 4.905 ( (V0 * sin[tex]\Theta[/tex]) / 9.81 )^2

I tried to solve for V0 and got sqrt(39.24/(sin[tex]\Theta[/tex])^2). After that I didn't know where to go or even if I was heading in the right direction. Any hint, nudge, push and/or shove is greatly appreciated.

-Brett

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# Homework Help: 2D Kinematics with multiple unknown variables.

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