A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.2 m/s at ground level. Its engines then fire and it accelerates upward at 4.20 m/s2 until it reaches an altitude of 960 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s2. (You will need to consider the motion while the engine is operating separate from the free-fall motion.)
(a) How long is the rocket in motion above the ground?
(b) What is its maximum altitude?
(c) What is its velocity just before it collides with the Earth?
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For A, I solved it such that it was the sqrt of the initial velocity squarted + 2(acceleration times final height of rocket) = 43.8 S, but that was incorrect. For B and C, I'm a bit lost as to what to do.
To solve this problem, let's break it down into two parts: the motion while the engine is operating and the free-fall motion.
First, let's focus on finding the time the rocket is in motion above the ground.
(a) How long is the rocket in motion above the ground?
The rocket initially has an upward velocity of 80.2 m/s. It undergoes constant acceleration of 4.20 m/s^2 until it reaches an altitude of 960 m. We can use the kinematic equation for motion with constant acceleration to find the time it takes to reach that altitude.
Using the equation: final velocity = initial velocity + (acceleration × time)
We have: 0 = 80.2 m/s + (4.20 m/s^2 × t)
Solving for t, we get: t = -80.2 m/s / (4.20 m/s^2) = -19.095 s
However, we need to consider only the positive time. So the rocket is in motion for a positive time of: t = 19.095 s
(b) What is its maximum altitude?
To find the maximum altitude, we need to determine at what point the rocket's velocity becomes zero. We can do this by using the kinematic equation for motion with constant acceleration.
Using the equation: final velocity^2 = initial velocity^2 + 2 × acceleration × displacement
Here, the initial velocity is 80.2 m/s, the acceleration is -9.80 m/s^2 (due to gravity), and the displacement is the maximum altitude, which we'll call H.
Substituting the values, we get: 0 = (80.2 m/s)^2 + 2 × (-9.80 m/s^2) × H
Solving for H, we have: H = [(80.2 m/s)^2] / [2 × 9.80 m/s^2] = 328.32 m
So, the rocket's maximum altitude is 328.32 m.
(c) What is its velocity just before it collides with the Earth?
Once the engine fails, the rocket goes into free fall with an acceleration of -9.80 m/s^2. To find the velocity just before collision, we need to find the time it takes to reach the ground from the maximum altitude.
Using the equation: final velocity = initial velocity + (acceleration × time)
Here, the initial velocity is 0 m/s, the acceleration is -9.80 m/s^2, and the displacement is the maximum altitude, H, which is 328.32 m.
Substituting the values, we get: 0 = 0 m/s + (-9.80 m/s^2) × t
Solving for t, we have: t = 2H / 9.80 m/s^2 = 2 × 328.32 m / 9.80 m/s^2 = 67 s
Now we can find the final velocity using the equation: final velocity = initial velocity + (acceleration × time)
Substituting the values, we get: final velocity = 0 m/s + (-9.80 m/s^2) × 67 s = -656.6 m/s
So, the velocity just before the rocket collides with the Earth is -656.6 m/s. Note that the negative sign indicates the direction is downward.
If you use
S=v0*t + (1/2)at²
you have
S=960m
v0=80.2 m/s
a=4.2 m/s²
Solve for t.
I get 9.6 sec. for going up, and 14 sec. for coming down (v0=0,a=-g,S=-960).
Check my work and post if you would like to check answers.