Amodel rocket is fired vertically upward from rest. its acceleration for the first three seconds is a(t) = 60t at which time the fuel is exhausted and it becomes a freely "falling" body. fourteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to -18 ft/s in 5 s. the rocket then "floats" to the ground at that rate.
at what time does the rocket reach its maximum height?
what is that height?
at what time does the rocket land?

(a) 8.4375 s

(b) 1409.06 ft

(c) 40.55 seconds

Explanation:

v’(t)=a(t)=60t therefore

For first three seconds

Since motion is vertical and we take upward direction as positive, acceleration is negative hence

a(t)=-32 and but since at t=0 then v=270 hence c=270

-32.2t+270=0 hence

t=270/32=8.4375 s

(b)

and since s(0)=270 then

(c)

v(15)=-32(14)+270=-178 ft/s

Average speed in the next 5 seconds becomes

The altitude is 914-490=424 ft

Time when rocket hits ground

Total time=3+14+23.6=40.55 seconds

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement