1. A catapult launches a test rocket vertically upward from a well, giving the r

Question

1. A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.8 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 1120 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 (in seconds) is the rocket in motion above the ground?

(b) What is its maximum altitude (in km)?

(c) What is its velocity just before it hits the ground(in m/s)?

Explanation / Answer

a) for 1120 m

1120 = ut + at^2 = 79.8t + 3.80t^2/2

1.9t^2 + 79.8t - 1120 = 0

solving t = 11.10 sec

after that .

it will atttain v = 79.8 + 3.80 x 11.10 = 121.98 m/s speed

so using h =ut + at^2/2

-1120 = 121.98t - 9.8t^2/2

4.9t^2 - 121.98t - 1120 = 0

t = 32.03 sec

total timre = 43.13 sec

b) after 1120 m

h = v^2/2g = 121.98^2/2g =759.14 m

total H =1879.14 m

c) v = u +at   = 121.98 - 9.81 x 32.03 = - 191.91 m/s

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.