Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 81.0 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 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?

Answer 1

here,

intial speed , u = 81 m/s

a = 3.9 m/s^2

h0 = 1020 m

let the maximum speed achived when accelrating be v1 and time taken be t1

using thrid equation of motion

v1^2 - u^2 = 2 * a * h0

v1^2 - 81^2 = 2 * 3.9 * 1020

solving for v1

v1 = 102.5 m/s

v1 = u + a * t1

102.5 = 81 + 3.9 * t1

solving for t1

t1 = 5.51 s

a)

for free fall period

let the time taken be t2

h0 = v1 * t2 - 0.5 * g * t2^2

- 1020 = 102.5 * t2 - 0.5 * 9.81 * t2^2

solving for t2

t2 = 28.3 s

the time period , t = t1 + t2 = 33.81 s

b)

the maximum height , hma = h0 + v1^2 /2g

hmax = 1020 + 102.5^2 /(2 * 9.81)

hmax = 1555.5 m

c)

the velocity before hitting the ground , v = sqrt(2*g * hmax)

v = sqrt(2*1555.5 * 9.81) m/s

v = 174.7 m/s