A rocket is fired vertically with an upward acceleration of 30 m/s2. After 20 s, the engine shuts off and the rocket then continues rising (while in free-fall). The rocket eventually stops rising and then falls back to the ground.
(a) What is the highest point the rocket reaches?
(b) What is the total time the rocket is in the air?
(c) What is the speed of the rocket just before it hits the ground?
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To solve this problem, we'll break it down into parts and use the kinematic equations of motion.
(a) To find the highest point the rocket reaches, we need to find the maximum height it reaches during its ascent.
We can use the equation:
vf^2 = vi^2 + 2ad
Where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance or displacement.
In this case, the rocket is fired vertically upward, so its initial velocity is zero. The acceleration is given as 30 m/s^2, and we need to find the displacement at the highest point.
First, we find the time it takes the rocket to reach the highest point by using the following equation:
vf = vi + at
Since the final velocity at the highest point is zero, we can say:
0 = 0 + 30t
Solving this equation, we find that t = 0 seconds.
So, it takes 0 seconds for the rocket to reach the highest point. Plugging this value of t into the equation, we find:
0 = 0 + 2ad
Simplifying the equation, we get:
0 = 2ad
Since the acceleration is positive, the displacement must be zero at the highest point. So, the rocket reaches a maximum height of zero. This means it does not go higher than the ground level.
(b) To find the total time the rocket is in the air, we need to consider the entire motion, including both the ascent and descent phases.
During the ascent phase, the rocket accelerates upward at 30 m/s^2 for 20 seconds. This means its velocity increases at a constant rate.
To find the displacement during this phase, we can use the equation:
d = vit + 1/2at^2
Since the initial velocity (vi) is zero, the equation simplifies to:
d = 0 + 1/2at^2
Plugging in the values, we get:
d = 1/2 * 30 * (20)^2 = 6000 meters
So, during the ascent phase, the rocket travels a vertical distance of 6000 meters.
During the descent phase, the rocket is under free-fall due to gravitational acceleration (g = 9.8 m/s^2). The time it takes for the rocket to reach the ground is the same as the time it took to reach the highest point.
Therefore, the total time the rocket is in the air is 20 seconds.
(c) To find the speed of the rocket just before it hits the ground, we can use the equation:
vf = vi + gt
Since the initial velocity (vi) is zero, the equation simplifies to:
vf = 9.8 * t
Plugging in the value of t as 20 seconds, we get:
vf = 9.8 * 20 = 196 m/s
So, the speed of the rocket just before it hits the ground is 196 m/s.
Start by computing the velocity and altitude achieved while the engine is on. Then write the equation for height and velocity during free fall. It's not that hard.
When the velocity is zero during the free fall stage, maximum height has been reached. Solve for that time, and the altitude reached then. So much for (a). For (b), solve for the time when the altitude is zero.
For (c), a shortcut would be to take the kinetic energy and altitude H at burnout, add 2gH to V^2 at burnout, and solve for the new V^2 at impact
Show us your work if you need further assistance here.