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?
you need to provide some units before attempting any calculations.
speaking of calculations, what have you gotten so far?
To find the maximum height the rocket will reach, we first need to determine the time it takes for the engines to fail. We can use the kinematic equation:
h = vi*t + (1/2)*a*t^2
Where:
h = height (560m)
vi = initial velocity (0 m/s, as the rocket starts from rest)
a = acceleration (2.25 m/s^2)
t = time taken to reach the given height
Rearranging the equation to solve for time (t):
t = sqrt((2*h)/a)
Plugging in the values:
t = sqrt((2*560)/2.25)
t ≈ sqrt(498.22)
t ≈ 22.32 seconds
Now, to find the maximum height the rocket will reach, we need to determine the distance it will travel after the engines fail. As the only force acting on the rocket is gravity, the rocket's upward acceleration will now be replaced by a downward acceleration of -9.8 m/s^2.
Using the kinematic equation:
h = vi*t + (1/2)*a*t^2
Where:
h = final height (unknown)
vi = initial velocity (0 m/s)
a = acceleration (-9.8 m/s^2)
t = time taken to fall (unknown, but we can use the same t as before)
Rearranging the equation to solve for h:
h = (1/2)*a*t^2
Plugging in the values:
h = (1/2)*(-9.8)*(22.32^2)
h ≈ -2465.5 m
Since height cannot be negative, this negative sign indicates that the rocket will crash below the launch pad. Therefore, the maximum height the rocket will reach is 560 m above the launch pad.
To find the time it takes for the rocket to crash back to the launch pad after the engines fail, we can use the kinematic equation:
h = vi*t + (1/2)*a*t^2
Where:
h = height (0 m, as the rocket is at the launch pad level)
vi = initial velocity (unknown)
a = acceleration (-9.8 m/s^2)
t = time taken to fall (unknown)
We can rearrange the equation to solve for time (t):
0 = vi*t + (1/2)*a*t^2
Simplifying the equation:
-4.9t^2 + vit = 0
Since we know the initial velocity (vi) is zero, the equation simplifies further:
-4.9t^2 = 0
Solving for time (t):
t ≈ 0 (one solution) or t ≈ ∞ (another solution)
The time taken to fall back to the launch pad, t, can be either 0 seconds (instantaneous crash) or never (if the rocket continues to ascend infinitely). Since the initial velocity is zero and the only force acting on it is gravity, the rocket will reach the launch pad with zero velocity. Hence, just before it crashes, the rocket will be moving with a speed of 0 m/s.