58. A catapult launches a test rocket vertically upward from a well, giving the
ID: 1731869 • Letter: 5
Question
58. A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.00 m/s2 until it reaches an altitude of 1000 m. At that point, its engines fail and the rocket goes into free fall, with an acceleration of 29.80 m/s2. (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? (You will need to consider the motion while the engine is operating and the free-fall motion separately.)
Explanation / Answer
a)
consider the motion upto 1000 m altitude :
Y = height gained before engines fail = 1000 m
vo = initial speed = 80 m/s
a = acceleration = 4 m/s2
t = time taken to reach 1000 m
using the equation
Y = vo t + (0.5) a t2
1000 = 80 t + (0.5) (4) t2
t = 10 sec
vf = final velocity of rocket at altitude of 1000 m
using the equation
vf = vo + a t
vf = 80 + (4 x 10)
vf = 120 m/s
consider the motion of rocket in free fall :
vo = initial velocity = 120 m/s
a = acceleration = - 29.8 m/s2
t' = time taken in free fall
Y = - 1000 m
using the equation
Y = vo t' + (0.5) a t'2
- 1000 = 120 t' + (0.5) (- 29.8) t'2
t' = 13.2 sec
total time in air = t + t' = 10 + 13.2 = 23.2 sec
b)
consider the motion of rocket in after engine fail upto maximum height:
vo = initial velocity = 120 m/s
a = acceleration = - 29.8 m/s2
vf = final speed at maximum height = 0 m/s
Yo = initial height = 1000 m
Yf = final height
using the equation
vf2 = vo2 + 2 a (Yf - Yo)
02 = 1202 + 2 (- 29.8) (Yf - 1000)
Yf = 1241.61 m
c)
consider the motion from heighest point to ground
vo = initial velocity = 0 m/s
a = acceleration = - 29.8 m/s2
vf = final speed just before hitting the ground = ?
Y = vertical displacement = - 1000 m
using the equation
vf2 = vo2 + 2 a Y
vf2 = 02 + 2 (- 29.8) (- 1000)
vf = 244.13 m/s
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