# Projectile Motion with vector functions (3D Motion in space velocity)

• Feb 19th 2009, 01:53 PM
acg716
Projectile Motion with vector functions (3D Motion in space velocity)
A gun is fired with angle of elevation 30°. What is the muzzle speed if the maximum height of the shell is 484 m? g=gravity (9.8)
What is the initial velocity?

My instructor gave us a hint:
v_0 is initial velocity.

At max height vertical velocity = 0. This gives v_0 sin(alpha) - gt = 0.

Vertical distance = -0.5 gt^2 + v_0 sin(alpha).

Find v_0 from the above two equation.

So I set those two equations equal to each other since they both equal zero. I'm not sure if I can do that. So my initial velocities cancelled out, I solved for t and got sqrt(2). I plugged that back into the vertical velocity and got v_0 = 28. I then plugged it back into the other equation to check it, but didn't get zero, I got 4.06.

I'm pretty lost.
• Feb 19th 2009, 02:30 PM
skeeter
$\displaystyle v_0\sin{\alpha} - gt = 0$

$\displaystyle t = \frac{v_0\sin{\alpha}}{g}$

$\displaystyle \Delta y = v_0\sin{\alpha} \cdot t - \frac{1}{2}gt^2$

$\displaystyle \Delta y = v_0\sin{\alpha} \cdot \frac{v_0\sin{\alpha}}{g} - \frac{1}{2}g\left(\frac{v_0\sin{\alpha}}{g}\right) ^2$

$\displaystyle \Delta y = \frac{v_0^2 \sin^2{\alpha}}{g} - \frac{v_0^2 \sin^2{\alpha}}{2g}$

$\displaystyle \Delta y = \frac{v_0^2 \sin^2{\alpha}}{2g}$

$\displaystyle v_0 = \sqrt{\frac{2g \Delta y}{\sin^2{\alpha}}}$