Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level. The engines then fire, and the rocket accelerates up- ward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point, its engines fail and the rocket goes into free fall, with an acceleration of 29.80 m/s2. (a) For what time inter- val is the rocket in motion above the ground?

Answer 1

There is an interval of 24.28s in which the rocket is above the ground.

Step-by-step explanation:

`y_(i)=0m`

`v_i=80(m)/(s)`

`a=4(m)/(s^2)`

`y_(e)=1000m`

`g=9.8(m)/(s^2)`

From Kinematics, the position
`y` as a function of time when the engine still works will be:

`y(t)=v_it+(1)/(2)at^2`

At what time the altitud will be
`y_(e)=1000m`?

`v_it+(1)/(2)at^2=y_(e)` ⇒
`(1)/(2)at^2+v_it-y_(e)=0`

Using the quadratic formula:
`t_1=10s`.

How much time does it take for the rocket to touch the ground? No the function of position is:

`y(t)=y_(e)+v_et-(1)/(2)gt^2`

Where our new initial position is
`y_(max)`, the velocity when the engine breaks is
`v_e=v_i+at=120(m)/(s)` and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

`y_(e)+v_et-(1)/(2)gt^2=0`

Again, using the quadratic ecuation:

`t_2=24.49s`

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

`t_T=t_1+t_2=34.49s`.