Question

A rocket is launched upward with a constant acceleration of 165 m/s^2. After 8.00 seconds of ascension a passenger on the rocket drops a rock out of it. How long does it take for this rock to hit the ground?

Answer 1

To solve this problem we will apply the concepts related to the linear kinematic movement. We will start by finding the speed of the body from time and the acceleration given.

Through the position equations we will calculate the distance traveled.

Finally, using this same position relationship and considering the previously found speed, we can determine the time to reach your goal.

For time (t) and acceleration (a) we have to,

$t = 8s, a = 165m/s^2$

The velocity would be,

$u = a*t \\u = 165*8\\u = 1320m/s$

Now the position is,

$h= \frac{1}{2} at^2$

$h = \frac{1}{2} 165*8^2$

$h = 5280m$

Now with the initial speed and position found we will have the time is,

$h=ut +\frac{1}{2} at^2$

$-5280=1320t - \frac{1}{2} 9.8t^2$

$4.9t^2-1320t-5280=0$

Solving the polynomian we have,

$t = 273.33s = 4.56minutes$

Therefore  the rocket will take to hit the ground around to 4.56min