Question

A test rocket is launched by accelerating it along a 200.0 m, 35 degree incline at 1.25 m/s/s starting

from rest. The instant the rocket leaves the incline, its engines turn off, it is subject only to gravity,

and air resistance can be ignored. Find

a) the maximum height above the ground the rocket reaches

Answer 1

Answer:

The maximum height above the ground the rocket reaches is 123.1 m.

Explanation:

Let's find the final velocity at a distance of 200 m:

$v_{f}^{2} = v_{0}^{2} + 2ad$

Where:

$v_{f}$ is the final speed =?

v₀ is the initial speed =0

a is the acceleration = 1.25 m/s²

d is the distance = 200 m

$v_{f} = \sqrt{2ad} = \sqrt{2*1.25 m/s{2}*200 m/s} = 22.4 m/s$    

Now, when the engines of the rocket turn off and it is subject only to gravity, the height reached is:

$v_{fy}^{2} = v_{0y}^{2} - 2gh$

Where:

$v_{f}$ = 0

$h = -\frac{v_{fy}^{2} - v_{0y}^{2}*sin(\theta)}{2g} = \frac{(22.4*sin(35))^{2}}{2*9.81 m/s^{2}} = 8.4 m$

Finally, the maximum height above the ground is:

$h_{max} = h + H$

Where H is the vertical component of the 200.0 meters.

$h_{max} = h + H = 8.4 m + 200.0 m*sin(35) = 123.1 m$

Therefore, the maximum height above the ground the rocket reaches is 123.1 m.

I hope it helps you!