Question

A model rocket is fired vertically upward from rest. Its acceleration for the first 3 seconds is a(t) = 60t at which time the fuel is exhausted and it becomes a freely "falling" body. 14 seconds later, the rocket's parachute opens, and the (down)?

velocity slows linearly to -18 ft/s in 5 s. The rocket then "floats" to the ground at that rate.

(b) At what time does the rocket reach its maximum height? (Give your answer correct to one decimal place.)

What is that height? (Give your answer correct to the nearest whole number.)

(c) At what time does the rocket land? (Give your answer correct to one decimal place.)

Answer 1

Given acceleration `a(t)=60t` for first 3 thee seconds, after starting from rest. Rewriting in terms of velocity `v`
`(dv(t))/dt=60t`
`=>dv(t)=60t cdot dt`
Integrating both sides we obtain
`v(t)=int 60t cdot dt`
`=>v(t)=60t^2/2 +C`
where `C` is constant of integration and is `0`, as per the initial condition, `v(0)=0`. We get
`v(t)=30t^2` .....1)
Now velocity at `t=3`, when all the fuel is exhausted is
`v(3)=30xx3^2=270ft s^-1`, assuming FPS system of units.

Rewriting (1) in terms of height
`(dh(t))/dt=30t^2`
`=>dh(t)=30t^2cdot dt`
To find out the height attained at `t=3`, we need to integrate the expression from `t=0` to `t=3`. We get
`h(3)=int_0^3 30t^2cdot dt`
`=>h(3)=| 30t^3/3|_0^3`
`=>h(3)=270ft`
`v(3) and h(3)` are initial conditions for the freely falling body.

At the maximum height `(h(3)+h_1)` velocity is zero. Acceleration due to gravity is in a direction opposite to the positive direction of motion. Applicable Kinematic equation is
`v=u+a t`
Inserting values we get
`0=270+(-32) t`
`=>t=270/32=8.4s`, rounded to one decimal place.
Rocket gains maximum height at `t=3+8.4=11.4s`

To calculate maximum height attained we use the kinematic relation
`v^2-u^2=2as`
`:.` `0^2-270^2=2xx(-32)h_1`
`=>h_1=270^2/64`
`=>h_1=270^2/64`
`=>h_1=270^2/64=1139.1`, rounded to one decimal place.
Maximum height attained is `h(3)+h_1=270+1139.1=1409.1ft`