A test rocket is fired vertically upward from a well. A catapult gives it an ini

Question

A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 79.2 m/s at ground level. Its engines then fire and it accelerates upward at 4.00 m/s2 until it reaches an altitude of 910 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s2. (You will need to consider the motion while the engine is operating separate from the free-fall motion.)
(a) How long is the rocket in motion above the ground?
s

(b) What is its maximum altitude?
km

(c) What is its velocity just before it collides with the Earth?
m/s

Explanation / Answer

Set up 2 equations for s(t): The first with the rocket firing and the 2nd with the rocket not firing. Assuming that at t=0 the rocket is at ground level you have s(0) = 0 b = 79.2 s''(t) = 2a --> 2a = 4.2 so a = 2.1 And the initial equation is s(t) = 2.1t^2 + 79.2t Now find t0 such that s(t0) = 1000 The 2nd equation has initial velocity of s'(t0) and the acceleration changes from 79.2 to -9.8 so a changes from 2.1 to -4.9 and initial position of 1000 The new equation is s1(t) = -4.9t^2 + s'(t0)t + 1000 Now calculate s1'(t1) = 0 and s1(t1) is the maximium height Calculate s1(t2) = 0 and t0 + t2 tells you how much time the rocket was above earth. Finally, s1'(t2) tells you the velocity at impact.

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