A model rocket is fired vertically upward with an initial velocity of 49 m/s from a tower 150 m high. (a) How long will it take the rocket to reach its maximum height? (b) What is the maximum height? (c) How long will it take for the rocket to pass its starting point on the way down? (d) What is the velocity when it passes the starting point on the way down? (e) How long will it take for the rocket to hit the ground?

 To solve these problems, we use the equations of motion under uniform acceleration due to gravity, which is roughly \( g = 9.8 \, m/s^2 \) downward.

(a) Time to reach maximum height:

We will use the equation \( v = u + at \), where - \( v \) is the final velocity (0 m/s at maximum height), - \( u \) is the initial velocity (49 m/s), - \( a \) is the acceleration due to gravity (-9.8 m/s^2, since it's upward motion we take it as negative), - \( t \) is the time taken to reach maximum height.

Setting the final velocity \( v \) at maximum height to 0 (as the rocket stops for a moment before descending), we get: \( 0 = 49 - 9.8t \)

Solving for \( t \), we get: \( t = 49 / 9.8 = 5 \) seconds.

(b) Maximum height:

The maximum height can be calculated using \( s = ut + (1/2)at^2 \), where - \( s \) is the displacement, - \( u \) is the initial velocity, - \( t \) is the time, - \( a \) is the acceleration due to gravity.

Using the time calculated in part (a): \( s = 49 \times 5 + (1/2)(-9.8)(5^2) \) \( s = 245 - (1/2)(9.8)(25) \) \( s = 245 - 122.5 \) \( s = 122.5 \) meters.

But we mustn't forget to add the height of the tower: Total maximum height from the ground = \( 150 + 122.5 = 272.5 \) meters.

(c) Time to pass the starting point on the way down:

For this part, we'll consider the entire journey up to the maximum height and back to the starting height. We know the time to reach the maximum height is 5 seconds, and since the motion is symmetric, it will take another 5 seconds to come back down to the height of the tower. So, the total time will be \( 5 + 5 = 10 \) seconds.

(d) Velocity when it passes the starting point on the way down:

The velocity at the starting point on the way down will be equal in magnitude but opposite in direction to the initial velocity (ignoring air resistance). Thus, the velocity will be \( -49 \, m/s \) (the negative sign indicates downward direction).

(e) Time for the rocket to hit the ground:

For this calculation, we'll use the time from the tallest point to the ground, because the trip from launch to the tallest point is already known (5 seconds). The starting velocity (u) here is 0 m/s at the maximum height, and we need to cover an additional 150 m downward.

Now use \( s = ut + (1/2)at^2 \). But this time, since the motion is downward, the acceleration is positive (9.8 m/s^2): \( 150 = (0)t + (1/2)(9.8)t^2 \) \( 150 = (1/2)(9.8)t^2 \) \( t^2 = 150 / (1/2)(9.8) \) \( t^2 = 150 / 4.9 \) \( t^2 = 30.61 \) \( t = \sqrt{30.61} \) \( t ≈ 5.53 \) seconds (approximately).

Adding the time to reach maximum height (5 seconds), total time from launch to hit the ground is \( 5 + 5.53 = 10.53 \) seconds.