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

Answer:

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

Explanation:

$y_{i}=0m$

$v_i=80\frac{m}{s}$

$a=4\frac{m}{s^2}$

$y_{e}=1000m$

$g=9.8\frac{m}{s^2}$

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

$y(t)=v_it+\frac{1}{2}at^2$

At what time the altitud will be $y_{e}=1000m$?

$v_it+\frac{1}{2}at^2=y_{e}$ ⇒ $\frac{1}{2}at^2+v_it-y_{e}=0$

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:

$y(t)=y_{e}+v_et-\frac{1}{2}gt^2$

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

Now, when the rocket tounches the ground:

$y_{e}+v_et-\frac{1}{2}gt^2=0$

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:

$t_T=t_1+t_2=34.49s$.