# Projectile Motion maximum altitude of rocket

A rocket is launched at an angle of 53o above the horizontal with an initial speed of 100m/s. It moves for 3s along its initial line of motion with an accelration of 30m/s2. At this time its engines fail and the rocket proceeds to move as a free body. Find a)the maximum altitude reached by the rocket, b)its total time of flight, and c)its horizontal range.

Can someone please explain how to approach this problem.

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tony873004
Gold Member
Start by figuring out its velocity after 3 seconds of accelerating.

This is what i have so far.

Part a:
s = vot + (.5)at^2
s = 100m/s(3s) + .5(30m/s/s)(3s)^2 = 435 m

y1 = s(sin 53) = 347 m
y2= [100(sin53)^2 / 2(9.8)] = 325 m
ytotal = 347 + 325 = 672 m

The rest of the problem will be solved incorrectly. But here is what i would have done if the previous answer was correct.

Part b:
t(total) = t1 + t2

t1= 3s
y= vo(sin53)t - .5*g*(t^2)
solve for time using quadratic eqn.

the correct answer is 36.1 s

Part c
x1 = v0t + .5(a)t^2 = 262m
x2 = s(cos 53) = 435m
xtotal = 435 + 262 = 697 m

the correct answer is 4050 m

tony873004