Rocket Launch problems?

Question

In physics we did a lab and shot rockets at different angles. I can't figure out these answers.

My only known information is the fifteen degree theta angle at which we shot it, and that the time was 1.77 seconds before it hit the ground. (also, it is known that when going up there is no acceleration, and going down the acceleration is -9.81 m/s/s)

I need:

1] intial vertical velocity at the time of launch, in m/s
2] max height reached in meters and feet
3] resultant velocity at which the rocket was launch in m/s and mi/h
4] horizontal velocity in m/s
and 5] horizontal range in meters and feet.

We have to use the kinematic equations, trig functions, and the pythag. theorem.

help?

Answer 1

The vertical velocity component is vy = v*sinø; the vertical ascent will continue until g*t = vy (there is acceleration on the ascent, it is just negative and subtracts from the initial velocity). The descent time is the same, so the total time is

T = 2*vy / g; so vy = .5*g*T

The max height is given by h = 0.5*g*T^2

If vy = v*sinø, then v = vy/sinø; use vy from above and ø= 15º

After you get v, the horizontal velocity is vx = v*cosø. (If you want to use the Pythagorean theorem, then v = √[vx^2 = vy^2]; solve for vx: vx = √[v^2 - vy^2].)

The rocket traveled at vx horizonally for the full ascent and descent time that you calculated above as T; therefore the horizontal range is T*vx = t*v*cosø