A 7500kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20m/s2 and feels no appreciable air resistance. When it has reached a height of 595m , its engines suddenly fail so that the only force acting on it is now gravity.
[1]What is the maximum height this rocket will reach above the launch pad?
[2]How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?
[3]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 first determine the time it takes for the rocket to reach that height using the initial conditions and kinematic equations.
1. Finding the time taken to reach 595m:
We can use the following kinematic equation to solve for time (t):
h = ut + (1/2)at^2
Where:
h = height (595m)
u = initial velocity (0 m/s since the rocket starts from rest)
a = acceleration (2.20 m/s^2)
t = time
Rearranging the equation to solve for t:
t^2 = (2h/a)
t^2 = (2 * 595m) / 2.20m/s^2
t^2 = 540.91s^2
t ≈ 23.29s
So, it takes approximately 23.29 seconds for the rocket to reach a height of 595m.
2. Finding the maximum height:
To find the maximum height (hmax) that the rocket will reach above the launch pad after the engines fail, we can use the following equation:
hmax = h + (ut - (1/2)at^2)
Since the rocket starts from rest initially, the equation simplifies to:
hmax = h - (1/2)at^2
Substituting the values we already know:
hmax = 595m - (1/2) * 2.20m/s^2 * (23.29s)^2
hmax ≈ 595m - 588.50m
hmax ≈ 6.50m
Therefore, the maximum height the rocket will reach above the launch pad after the engines fail is approximately 6.50 meters.
3. Finding the time before the rocket crashes back down:
When the engines fail, the only force acting on the rocket is gravity, causing it to accelerate downward with a value of -9.81 m/s^2. To find the time it takes for the rocket to crash back to the launch pad, we can use:
h = ut + (1/2)at^2
Where:
h = initial height (595m - 6.50m = 588.50m)
u = initial velocity (0 m/s since the rocket starts from rest after the engine failure)
a = acceleration (gravity = -9.81 m/s^2)
t = time
Rearranging the equation to solve for t:
t^2 = (2h/a)
t^2 = (2 * 588.50m) / 9.81m/s^2
t^2 = 120.10s^2
t ≈ 10.96s
Therefore, it takes approximately 10.96 seconds for the rocket to crash back down to the launch pad after the engines fail.
4. Finding the speed just before crashing:
To find the speed just before the rocket crashes back down to the launch pad, we can use the equation:
v = u + at
Where:
v = final velocity (what we are looking for)
u = initial velocity (0 m/s since the rocket starts from rest after the engine failure)
a = acceleration (gravity = -9.81 m/s^2)
t = time (10.96s)
Substituting the values:
v = 0 m/s + (-9.81 m/s^2) * 10.96s
v ≈ -107.5 m/s
Since the velocity is negative, we can disregard the direction and take the magnitude. Therefore, the speed just before the rocket crashes back down to the launch pad is approximately 107.5 m/s.