A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 60

Question

A model 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.

a) At what time does the rocket reach its maximum height?

What is that height?

b) At what time does the rocket land?

Answer 1

Let's calculate height and velocity at the end of the engine burn first:

v(t)=.5*60*t^2
v(3)=30*9
v(3)=270
(Is this ft/s? I will assume since those units are used next)

y(t)=30*t^3/3
y(3)=10*27
y(3)=270 ft

for the next part of the flight the rocket has
v(t)=270-32*t
to find apogee, set v(t)=0
t=270/32
t=8.4375 seconds
This is 11.4375 into the flight

y(t)=270+270*t-16*t^2
y(8.4375) is apogee
=1,409 feet

b)when the parachute opens

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

and the altitude is
y(14)=270+270*14-16*14^2
=914 ft

over the next 5 seconds, the average speed
is -(178+18)/2
=-98 ft/s

and it achieves a constant downward speed
of -18 ft/s

during the 5 seconds, the distance covered
is 98*5
490 feet, so the altitude is
914 - 490
or 424 ft
to traverse 424 ft at 18 ft/s
23.6 seconds
so the total flight time is
3+14+23.6
or 40.6 seconds when the rocket touches down.

j

Answer 2

the final one is 45.56 or 45.6