A 7600 rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 and feels no appreciable air resistance. When it has reached a height of 560 , its engines suddenly fail so that the only force acting on it is now gravity.What is the maximum height this rocket will reach above the launch pad? How much time after engine failure will elapse before the rocket comes crashing down to the launch pad? How fast will it be moving just before it crashes?

To find the maximum height the rocket will reach above the launch pad, we can use the equation for displacement:

s = ut + (1/2)at^2

Where:
s = displacement (height)
u = initial velocity (0 m/s as it starts from rest)
a = acceleration (2.25 m/s^2)
t = time

We need to find the time it takes for the rocket to reach a height of 560 m. Rearranging the equation, we have:

t = sqrt((2s) / a)

Substituting the values, we get:

t = sqrt((2 * 560) / 2.25) = sqrt(1120 / 2.25) = sqrt(498.22) ≈ 22.34 s

Therefore, the time it takes for the rocket to reach a height of 560 m is approximately 22.34 seconds.

Next, to calculate the maximum height, we substitute the time value into the formula for displacement:

s = ut + (1/2)at^2

s = 0 * 22.34 + (1/2) * 2.25 * (22.34)^2

s ≈ 2498.02 m

Therefore, the maximum height the rocket will reach above the launch pad is approximately 2498.02 meters.

Now, to determine the time it takes for the rocket to come crashing down to the launch pad, we use the equation:

s = ut + (1/2)at^2

Since the rocket started at a height of 560 m and reached a maximum height of 2498.02 m, the total displacement is:

total displacement = 2498.02 - 560 = 1938.02 m

Since the rocket is moving upwards initially and then falls back down, the displacement during the descending portion is equal to the negative of the total displacement:

descending displacement = -1938.02 m

Now, we can find the time it takes for the rocket to fall back to the launch pad by solving for t in the equation:

-1938.02 = 0 * t + (1/2) * (-9.8) * t^2

Simplifying the equation:

-1938.02 = -4.9t^2

Rearranging the equation:

t^2 = 1938.02 / 4.9

t^2 ≈ 395.12

Taking the square root:

t ≈ sqrt(395.12) ≈ 19.88 s

Therefore, it will take approximately 19.88 seconds for the rocket to come crashing down to the launch pad.

Finally, to calculate the velocity just before it crashes, we can use the equation:

v = u + at

Since the rocket's engines failed, the initial velocity is 0 m/s. The acceleration due to gravity is -9.8 m/s^2 (negative sign because it is acting downwards).

v = 0 + (-9.8) * 19.88

v ≈ -194.02 m/s

Therefore, the rocket will be moving at approximately -194.02 m/s (downwards) just before it crashes to the launch pad.