Parabolic Rocket Flight Path Challenge Question?

Question

Given:
Angle of Launch is Θ; 0° < Θ < 90°
Velocity of Launch is V
For simplicity, gravity causes objects to fall at 5t² meters, where t is the number of seconds the object has fallen.
No air/wind effects
Landing location is the same elevation as launch location on a flat plane.
Flight path is in 2 dimensions.

Express in terms of V and Θ:

Total number of seconds the rocket is in flight
Total number of meters the rocket travels horizontally
Total number of meters above ground the rocket reaches at its highest point

Thanks for your consideration!

Perhaps Dr. D is suggesting it's not really a challenge question...Well, Okay. I already know the answers, and I'm challenging my precalculus students to produce them, and I wanted to see if Yahoo! Answerers thought it as challenging as they. Apparently you don't. In that context, maybe your answers will be different. Thanks!

Answer 1

The first step is to determine total flight time. The rocket will reach apogee at exactly 1/2 the total flight time.

using v(t)=v*sin(Θ)-10*t
set v(t)=0 and solve for t

t=v*sin(Θ)/10

soo the total flight time is
2*v*sin(Θ)/10

Horizontal motion is
x(t)=v*cos(Θ)*t
plug in th eflight time for t
x(range)=v*cos(Θ)*2*v*sin(Θ)/10
simplify
x(range)=v^2*2*sin(2*Θ)/10

y(t)=v*sin(Θ)*t-5*t^2
to find the apogee, plug in t at apogee
v*sin(Θ)/10
y(apogee)=v^2*sin^2(Θ)/10-5*v^2*sin^2(Θ)/100
simplify
y(apogee)=v^2*sin^2(Θ)/20
j

Answer 2

This is a standard derivation which you'll find in any textbook on projectile motion.

Horizontal motion:
u = V*cosθ
x = (V*cosθ)*t
t = x / V*cosθ

Vertical motion:
v = V*sinθ - gt
y = (V*sinθ)*t - gt^2 / 2
= (V*sinθ)*x / V*cosθ - g*(x / V*cosθ)^2 / 2
= x*tanθ - (5*sec^2 θ/V^2) * x^2