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

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:

Where:

is the final speed =?

v₀ is the initial speed =0

a is the acceleration = 1.25 m/s²

d is the distance = 200 m

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

Where:

= 0

Finally, the maximum height above the ground is:

Where H is the vertical component of the 200.0 meters.

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

I hope it helps you!                                                          

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force