A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.8 m/s at ground level. Its engines then fire and it accelerates upward at 3.90 m/s2 until it reaches an altitude of 920 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?
__________s
(b) What is its maximum altitude?
_______km
(c) What is its velocity just before it collides with the Earth?
______m/s
Please explain this question and show what equations should be used ! Thanks!
You have three motions to consider.
Phase 1: powered by engine
With initial velocity of v0=80.8 m/s
it slows down due to gravity until a height of S=920 m. and an acceleration of a=3.9 m/s².
THe governing equation is
S = v0*t + (1/2)a*t²
Solve for t.
The velocity at the end of this phase is
v1=v0+a*t
Phase 2: deceleration until maximum height is reached, at which time velocity = 0.
Time to reach maximum height
t1 = v1/g
Additional height travelled:
S1=v1*t1 +(1/2)(-g)t1²
Phase 3: Free fall from maximum height S+S1
Use the usual equations to calculate the velocity and time
To solve this problem, we can break it down into two parts: the motion of the rocket while the engine is operating and the free-fall motion after the engines fail. Let's solve it step by step:
(a) Motion while the engine is operating:
We can use the kinematic equation to find the time taken to reach the altitude of 920 m. The equation we need is:
h = ut + (1/2)gt^2
where h is the vertical displacement, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
Given:
u = 80.8 m/s (initial speed)
g = 9.80 m/s^2 (acceleration due to gravity)
Setting the displacement h to 920 m and solving for t, we get:
920 = 80.8t + (1/2)(3.90)t^2
Simplifying and rearranging the equation:
0.195t^2 + 80.8t - 920 = 0
We can solve this quadratic equation to find the time taken to reach an altitude of 920 m. Using the quadratic formula, we get:
t = (-80.8 ± √(80.8^2 - 4(0.195)(-920))) / (2(0.195))
Solving this equation, we find two possible solutions for t:
t1 ≈ 7.51 s
t2 ≈ -48.26 s (this negative value is not valid)
Therefore, the rocket is in motion above the ground for approximately 7.51 seconds.
(b) Maximum altitude:
To find the maximum altitude, we need to calculate the displacement at that point using the equation:
h = ut + (1/2)gt^2
Using the given values:
u = 80.8 m/s (initial speed)
g = 9.80 m/s^2 (acceleration due to gravity)
t = 7.51 s (time when the rocket reaches its maximum altitude)
Substituting these values into the equation, we have:
h = (80.8)(7.51) + (1/2)(-9.80)(7.51)^2
Simplifying this equation, we find:
h ≈ 283.8 m
Therefore, the maximum altitude of the rocket is approximately 283.8 meters.
(c) Velocity just before colliding with the Earth:
After the engines fail, the rocket goes into free fall with an acceleration of -9.80 m/s^2. To find the velocity just before it collides with the Earth, we can use the equation:
v = u + gt
We know:
u = 0 m/s (as the rocket is in free fall)
g = -9.80 m/s^2 (acceleration due to gravity)
t = 7.51 s (time when the rocket reaches the maximum altitude)
Substituting these values into the equation, we have:
v = 0 + (-9.80)(7.51)
Simplifying this equation, we find:
v ≈ -73.49 m/s
Therefore, the velocity just before the rocket collides with the Earth is approximately -73.49 m/s (with the negative sign indicating the velocity is directed downward).
To solve this problem, we need to consider the two phases of motion: when the engines are operating (first phase) and when the rocket goes into free fall (second phase).
(a) The first phase of motion is when the rocket is accelerated by its engines. We can use the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 80.8 m/s (initial velocity)
a = 3.90 m/s^2 (acceleration)
To find the time it takes for the rocket to reach an altitude of 920 m, we can rearrange the equation:
t = (v - u) / a
Since the final velocity at this altitude is not given, we will use the equation:
v^2 = u^2 + 2ad
where:
d = displacement
Rearranging this equation to solve for displacement, we get:
d = (v^2 - u^2) / (2a)
Given:
d = 920 m
u = 80.8 m/s
a = 3.90 m/s^2
Substituting the given values into the equation, we can find the final velocity at an altitude of 920 m.
Once we have the final velocity, we can calculate the time it took for the rocket to reach this altitude using the first equation.
(b) To find the maximum altitude, we need to find the displacement at maximum height. The displacement is the distance traveled from the ground to the maximum altitude, and in this case, it is equal to 920 m.
(c) To find the velocity just before the rocket collides with the Earth, we need to consider the second phase of motion when the rocket goes into free fall. In this phase, the rocket's acceleration is -9.80 m/s^2 (negative due to the direction of the acceleration being downward).
In free fall motion, we can use the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = final velocity at maximum altitude (calculated in part (a))
a = -9.80 m/s^2 (acceleration)
We can use this equation to find the velocity just before the rocket collides with the Earth.
By solving these equations, we will be able to find the answers to all three parts of the question.