Question

A stationary rocket is launched vertically upward. After 4s, the rocket's fuel is used up and it is 225.6 m above the ground. At this instant the velocity of the rocket is 112,8 m/s. Then the rocket undergoes free fall. Ignore the effect of the air friction.Assume that g does nt change the entire motion of the rocket. Taking up as positive,use equation of motion to determine the time taken from the moment the rocket is launched until it strikesthe ground.

Answer 1

Tr = To + (-Vo)/g = 4 + (-112.8)/-9.8 =
15.51 s. = Rise time.

hmax = ho + (V^2-Vo^2)/2g
hmax = 225.6 + (0-(112.8)^2)/-19.6 =
1506 m. Above gnd.

hmax = 0.5g*t^2 = 1506 m.
4.9t^2 = 1506
t^2 = 307.3
Tf = 17.53 s. = Fall time.

T = Tr + Tf = 15.51 + 17.53 = 33.0 s. =
Time in air.