recall that the equation of motion is
y = h0 + tanθ x - g/(2v^2) sec^2θ x^2
so,
y = 200 + 1.11x - 0.00001787x^2
y=0 at x=62278.7 ft or 11.8 miles
A rocket is launched from a tower that is 200 feet high at 1000ft/sec at 48°.
How far does the rocket travel?
1 answer