2D Kinematics with multiple unknown variables.

[Blbuford]

Homework Statement

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.

Homework Equations

y - y0 = Vy0 * t + 1/2 * ay * t^2
Vy = Vy0 + ay * t
x - x0 = Vx0 * t

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

 

[ehild]

Just a piece of advice: do not substitute the numerical data to soon. Keep using the symbol g during the derivations. Also, you can use v_x0 instead of v₀cosθ and v_y0 instead of v₀sinθ.

So you can get v_y0=v₀sinθ from your last equation. Plug-in this value for v_y0 into the original equation for v_y0. You will find the total time of flight by using the given data: y0=0 and y(final)=-1 m.

ehild