posted by ann on .
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?
(b) What is its maximum altitude in kilometers?
(c) What is its velocity just before it hits the ground?
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.