A catapult launches a test rocket vertically upward from a well, giving the rock

Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.80 m/s2 until it reaches an altitude of 1020 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 and the free-fall motion separately.)

(a) For what time interval is the rocket in motion above the ground?

(b) What is its maximum altitude?

(c) What is its velocity just before it hits the ground?

Explanation / Answer

Here ,

a) let the time interval is t

Using second equation of motion

d = u*t + 0.5 * a*t^2

1020 = 80.6 * t + 0.5 * 3.80 * t^2

solving for t

t = 10.20 s

the time interval for which the rocket is in air is 10.20 s

b)

it speed at 1020 m

v = sqrt(2 * 3.80 * 1020 + 80.6^2)

v = 119.4 m/s

maximum height = v^2/(2 * g) + 1020

maximum height = 119.4^2/(2 * 9.8) + 1020

maximum height = 1747 m

the maximum height of rocket is 1747 m

c)

let the velocity is v

v = sqrt(2 * 9.8 * 1747)

v = 185 m/s

the speed of rocket just befoe hitting the ground is 185 m/s

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