Answer:
625 ft
Step-by-step explanation:
We assume ballistic motion, which would be a bit unusual for a rocket.
The horizontal distance traveled in time t is ...
x = 200·cos(15°)·t
The vertical position at time t is ...
y = -16t^2 +200·sin(15°)t
Substituting for t using the first equation, this becomes ...
y = -16(x/(200·cos(15°)))^2 +200·sin(15°)(x/(200·cos(15°)))
Setting y=0 and factoring out x, we get ...
0 = x(-16x/(40000cos(15°)^2) +sin(15°)/cos(15°))
Multiplying by 2cos(15°)^2/x, this becomes ...
0 = -32x/40000 +2sin(15°)cos(15°)
or ...
sin(30°) = x(32/40000) = x/1250
1250·sin(30°) = x = 625 . . . . . feet
The rocket will travel 625 feet.
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Comment on the general solution
To summarize the general result here,
x = v^2·sin(2θ)/32
where x is the distance traveled, v is the initial velocity, and θ is the launch angle. The trajectory is assumed to be from ground level to ground level.
