Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.0 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 980 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?
[40.51 seconds]

(b) What is its maximum altitude in kilometers?
[? km]

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

Answer 1

V^2 = Vo^2 + 2a*h
V^2 = 79^2 + 7.8*980 = 13,885
V = 117.83 m/s. @ 980 m

a. V = Vo + a*t = 117.83
79 + 3.9t = 117.83
3.9t = 117.83-79 = 38.83
Tr1 = 9.96 s. to rise to 980 m.

V = Vo + g*t
117.83 - 9.8t = 0
9.8t = 117.83
Tr2 = 12.02 s to reach max. ht.

hmax = ho + (V^2-Vo^2)/2g
hmax=980 + (0-(117.83)^2)/-19.6=1688 m.
= max. ht

hmax = Vo*t + 0.5g*t^2 = 1688 m
0 + 4.9t^2 = 1688
t^2 = 345
Tf = 18.57 s. = Fall time.

T = Tr1 + Tr2 + Tf
T = 9.96+12.02+18.57 = 40.55 Seconds.

b. hmax = 1688 m. = 1.688 km. See previous calculation.

c. V^2 = Vo^2 + 2g*hmax
V^2 = 0 + 19.6*1688 = 33,084.8
V = 182 m/s.