a 7500 kilogram rocket blasts off vertically upward from the launch pad with a constant acceleration of 2.25 m/s squared and feels no appreciable air resistance, when it reaches the height of 525 meter its engine suddenly fail so that the only force acting on it is now gravity. (a.) what is the maximum height the rocket will reach above the launch pad? (b.) how much time after engine failure will elapse before the rocket comes crashing down and how fast will it be moving just before it crashes?
For simplicity, assume g = constant over the trajectory. This assumes the height change is a small fraction of the earth's radius, 6000 km.
(a) First get the height (H' = 525 m) and time T at engine cutoff.
a*T^2/2 = 525
T = 21.6 s
Velocity at cutoff = Vm = a T = 48.6 m/s
Max height Hm = H' + additional height gained while "coasting" upwards:
Hm = 525 m + Vm^2/(2g)= 645.5 m
(b) Let additional time T'to "crash".
Solve g*T' = Vm, for T'.
To compute speed V" at impact:
V"^2/2 = Vm^2/2 + gH'
To solve this problem, we will use the equations of motion and consider two parts of the rocket's motion: when the engine is functioning and when it fails.
(a). To determine the maximum height the rocket reaches above the launch pad, we can use the first part of its motion with constant acceleration.
The initial velocity (u) is 0 m/s since the rocket starts from rest, and the acceleration (a) is 2.25 m/s^2. The distance traveled (s) is 525 meters.
We can use the equation:
s = ut + (1/2)at^2
Plugging in the values, we have:
525 = 0*t + (1/2)*(2.25)*t^2
Simplifying the equation, we get:
525 = (1.125)t^2
Dividing both sides of the equation by 1.125, we have:
t^2 = 525/1.125
t^2 = 466.67
Taking the square root of both sides, we find:
t ≈ 21.58 seconds
Thus, the time it takes for the rocket to reach the maximum height is approximately 21.58 seconds.
To find the maximum height (h), we can substitute the value of t into the equation:
h = ut + (1/2)at^2
h = 0*t + (1/2)*(2.25)*t^2
Plugging in the value of t, we get:
h = 0 + (1/2)*(2.25)*(21.58)^2
Simplifying the equation, we find:
h ≈ 2727.5 meters
Therefore, the maximum height the rocket reaches above the launch pad is approximately 2727.5 meters.
(b). After the engine failure, the only force acting on the rocket is gravity (g). The acceleration due to gravity is 9.8 m/s^2.
To find the time it takes for the rocket to crash back to the ground, we assume it starts from the maximum height reached and travels downward to the ground.
Using the equation:
s = ut + (1/2)at^2
We plug in the values:
-2727.5 = 0*t + (1/2)*(-9.8)*t^2
Simplifying the equation, we have:
-2727.5 = -(4.9)t^2
Dividing both sides of the equation by -1, we find:
2727.5 = 4.9*t^2
Dividing both sides by 4.9, we get:
556.12 ≈ t^2
Taking the square root of both sides, we find:
t ≈ 23.57 seconds
Therefore, the time it takes for the rocket to crash after engine failure is approximately 23.57 seconds.
To find the speed of the rocket just before it crashes, we can use the equation:
v = u + at
As the rocket is moving downward, the initial velocity (u) is 0 m/s, the acceleration (a) is the acceleration due to gravity (9.8 m/s^2), and the time (t) is approximately 23.57 seconds.
Plugging in the values, we have:
v = 0 + (9.8)*(23.57)
Calculating the value, we find:
v ≈ 230.53 m/s
Therefore, the rocket will be moving at approximately 230.53 m/s just before it crashes.
To answer these questions, we can use the kinematic equations of motion. Here's how we can find the answers step by step:
(a.) What is the maximum height the rocket will reach above the launch pad?
To find the maximum height, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (which is zero when the rocket reaches its maximum height)
vi = initial velocity
a = acceleration
d = displacement or height
Rearranging the equation, we get:
d = (vf^2 - vi^2) / (2a)
Since the rocket initially moves upward with a constant acceleration, its final velocity when it reaches the maximum height will be zero. Therefore, we can simplify the equation to:
d = -vi^2 / (2a)
Given:
vi = 0 (since the rocket starts from rest at the launch pad)
a = 2.25 m/s^2
d = 525 meters
Plugging in the values, we can calculate the maximum height:
d = -0^2 / (2 * 2.25)
d = 0 meters
Therefore, the maximum height the rocket will reach above the launch pad is 525 meters.
(b.) How much time after engine failure will elapse before the rocket comes crashing down, and how fast will it be moving just before it crashes?
To find the time elapsed before the rocket comes crashing down, we need to find the time it takes for the rocket to go from the maximum height to the ground. We can use the equation:
d = vi * t + (1/2) * a * t^2
Where:
d = displacement (which is the maximum height - 525 meters)
vi = initial velocity (which is zero when the rocket reaches its maximum height)
a = acceleration due to gravity (-9.8 m/s^2, since the rocket is now only affected by gravity)
t = time elapsed (what we want to find)
Rearranging the equation, we get a quadratic equation:
0.5 * (-9.8) * t^2 + 0 = -d
Simplifying the equation further, we get:
4.9 * t^2 = d
Plugging in the values, we can solve for t:
4.9 * t^2 = 525
t^2 = 525 / 4.9
t^2 = 107.14
t = √107.14
t ≈ 10.35 seconds
Therefore, it will take approximately 10.35 seconds after the engine failure for the rocket to come crashing down.
To find the velocity just before the rocket crashes, we can use the equation:
vf = vi + at
Where:
vi = initial velocity (which is zero when the rocket reaches its maximum height)
a = acceleration due to gravity (-9.8 m/s^2)
vf = final velocity (what we want to find)
Plugging in the values, we can solve for vf:
vf = 0 + (-9.8) * 10.35
vf ≈ -101.43 m/s
The negative sign indicates that the velocity is in the downward direction.
Therefore, just before the rocket crashes, it will be moving at approximately -101.43 m/s.