A model rocket is launched straight upward with an initial speed of 50 m/s. It accelerates with a constant upward acceleration of 2.00 m/s^2 until its engines stop at an altitude of 150 m.

a) What is the max. height reached by the rocket?
b) When does the rocket reach max. height?
c) How long is the rocket in the air?

I really need help with which equation to use for this problem. I'm unsure of even where to begin...!

a) First, calculate the velocity Vmax attained during acceleration at rate a. (They already tell you the altitude there, H = 150 m). Then compute how huch higher it "coasts" before reaching maximum altitude.

Vmax = sqrt(2aH)= 24.5 m/s
To have the velocity decrease to zero, the additional time of flight t' is given by
g t' = 24.5 m/s, so
t' = 24.5/9.8 = 2.5 s
The average speed while coasting to zero velocity is Vmax/2 = 12.25 m/s.
Additionl altitude gained = (2.5)(12.25) = 30.63 m
Maximum altitude = Hmax = 150 + 30.63 = 180.63 m

b) it attains maximu height 2.5 seconds after accelerqation stopped. We already calculated that. Add that to the tiome spent accelerating for the total time after launch. The time t spent accelerating is given by
(1/2) a t^2 = 150 m
t = sqrt (2*150/a)= 12.25 s
Total time to reach maximum altitude - t + t' = ?

c) Add to the last answer the time that it takes to fall back down. Call this time t"

(1/2) g t"^2 = Hmax

Solve for t" and add it to the time in the previous answer

For part a, my book says the answer is 310 m. I'm still really confused on how they got this answer.

To solve this problem, we can use the equations of motion to find the maximum height reached by the rocket, the time it takes to reach the maximum height, and the total time the rocket is in the air.

Let's break it down step-by-step:

Step 1: Identify the given values and what you need to find.
Given:
- Initial velocity (u) = 50 m/s (upward)
- Acceleration (a) = 2.00 m/s^2 (upward)
- Altitude at engine stop (h) = 150 m

To Find:
a) Maximum height reached by the rocket
b) Time taken to reach the maximum height
c) Total time the rocket is in the air

Step 2: Choose the appropriate equation(s) to use.
To solve for height, we can use the kinematic equation:

v^2 = u^2 + 2as

Since we know the initial velocity, final velocity, acceleration, and height, we can rearrange the equation to solve for height:

h = (v^2 - u^2) / (2a)

To solve for time, we can use another kinematic equation:

v = u + at

Since we know the initial velocity, acceleration, and time, we can rearrange the equation to solve for time:

t = (v - u) / a

Step 3: Solve each part of the problem.
a) To find the maximum height reached by the rocket, we need to calculate the final velocity (v).
Using the second kinematic equation:

v = u + at

Substituting the given values:
u = 50 m/s
a = 2.00 m/s^2
t = ? (unknown)

Since we are looking for the maximum height, at this point, the velocity becomes zero because the engines stop.
So, v = 0 m/s.

Rearranging the equation to solve for time:
0 = 50 + 2t
2t = -50
t = -25 s

The negative time value does not make sense in this context, so we discard it.

b) To find the time it takes to reach the maximum height, we can use the equation:

t = (v - u) / a

Substituting the values:
u = 50 m/s
a = 2.00 m/s^2
v = 0 m/s (at maximum height)

t = (0 - 50) / 2
t = -25 s

Again, the negative time value doesn't make sense, so we discard it.

c) Finally, to find the total time the rocket is in the air, we need to find the time it takes to reach the maximum height and double it.

Total time = 2 * time to reach the maximum height

Using the last obtained valid time value:
Total time = 2 * 25
Total time = 50 s

Therefore:
a) The maximum height reached by the rocket is 150 m.
b) The rocket reaches the maximum height at 25 seconds.
c) The rocket is in the air for a total of 50 seconds.

Remember, it is essential to discard any non-physical negative values in the context of the problem.

To solve this problem, you can use the equations of motion, specifically the kinematic equations. The kinematic equations relate the displacement (Δy), initial velocity (Vi), final velocity (Vf), acceleration (a), and time (t). The kinematic equation that involves these variables is:

Δy = Vi * t + (1/2) * a * t^2

Now, let's tackle each part of the problem:

a) What is the max. height reached by the rocket?

To find the maximum height, we need to calculate the displacement (Δy) when the rocket's velocity (Vf) becomes zero. Since the acceleration is acting in the opposite direction of the rocket's motion, it will eventually decrease the velocity to zero.

Using the kinematic equation: Vf = Vi + a * t

We can solve for time (t) when the rocket's velocity becomes zero: 0 = 50 + (-2) * t

Solving for t: t = 50 / 2 = 25 seconds

Now, substitute the value of t into the displacement equation:

Δy = Vi * t + (1/2) * a * t^2
= 50 * 25 + (1/2) * (-2) * (25)^2
= 1250 - 625
= 625 meters

So, the maximum height reached by the rocket is 625 meters.

b) When does the rocket reach the max. height?

From the calculations above, we determined that the time (t) when the rocket reaches the maximum height is 25 seconds.

c) How long is the rocket in the air?

To find the total time the rocket is in the air, we can sum up the time when the rocket was ascending until it reached the maximum height (t1) and the time it takes for the rocket to descend back to the ground (t2).

Time taken to reach max. height: t1 = 25 seconds

Now, to find the time it takes for the rocket to descend, we can use the equation: Vf = Vi + a * t

Since the rocket is now descending, the acceleration is due to gravity and is equal to -9.8 m/s^2.

0 = 0 + (-9.8) * t2

Solving for t2: t2 = 0 / (-9.8) = 0 seconds

Since the rocket descends back to the ground from the maximum height instantaneously (t2 = 0 seconds), the total time the rocket is in the air is:

Total time = Time to reach max. height (t1) + Time to descend (t2)
= 25 + 0
= 25 seconds

Therefore, the rocket is in the air for 25 seconds.

I hope this helps you understand how to approach and solve this problem using the relevant equations!