A rocket is fired at a speed of 91.0 m/s from ground level, at an angle of 49.0 ° above the horizontal. The rocket is fired toward an 21.4-m high wall, which is located 25.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

Well, it looks like this rocket is a real overachiever! It wants to clear the top of the wall, huh? Let's see if it has what it takes.

First, let's break down the rocket's initial velocity into horizontal and vertical components. The horizontal component can be calculated using some trigonometry. We have a speed of 91.0 m/s at an angle of 49.0°. So, the horizontal component of the velocity is 91.0 m/s * cos(49.0°).

Using some math magic, we find that the horizontal component is approximately 58.437 m/s.

Now, let's focus on the vertical component. The vertical component can be calculated using the same trigonometry, but this time using the sine function. So, the vertical component of the velocity is 91.0 m/s * sin(49.0°).

After crunching the numbers, we find that the vertical component is approximately 68.406 m/s.

Okay, now we know the rocket's initial vertical velocity. Let's find out how much time it takes for the rocket to reach its maximum height. We can use the fact that the vertical displacement at the maximum height is zero.

Using the kinematic equation: vf = vi + at, where vf is the final vertical velocity, vi is the initial vertical velocity, a is the acceleration, and t is the time, we can set vf = 0 and solve for t.

We know that the acceleration is -9.8 m/s² (due to gravity), and the initial vertical velocity is 68.406 m/s.

Plugging in the values and doing some more math magic, we find that the time it takes to reach the maximum height is approximately 6.971 seconds.

Now, let's find out how high the rocket goes. We can use the kinematic equation: vf^2 = vi^2 + 2ad, where vf is the final vertical velocity, vi is the initial vertical velocity, a is the acceleration, and d is the displacement.

We know that the final vertical velocity is 0 m/s (at the maximum height), the initial vertical velocity is 68.406 m/s, and the acceleration is -9.8 m/s². We want to find the displacement (how high the rocket goes).

Plugging in the values, we find that the displacement is approximately 236.61 meters.

Finally, to find out how much the rocket clears the top of the wall, we subtract the height of the wall (21.4 meters) from the displacement of the rocket (236.61 meters).

After some last-minute calculations, we find that the rocket clears the top of the wall by approximately 215.21 meters.

So, this rocket is a real high flyer! It clears the top of the wall with ease. I hope it doesn't get a big head from all this success!

To find how much the rocket clears the top of the wall, we need to calculate the maximum height reached by the rocket's trajectory. Let's break down the problem into different parts.

Step 1: Find the vertical component of the initial velocity:
The vertical component of the initial velocity can be calculated using the formula:
Vyi = V * sin(θ)
where Vyi is the vertical component of the initial velocity, V is the launch speed (91.0 m/s), and θ is the launch angle (49.0°).

Vyi = 91.0 m/s * sin(49.0°)
Vyi = 91.0 m/s * 0.7547
Vyi ≈ 68.70 m/s

Step 2: Find the time it takes for the rocket to reach the maximum height:
To find the time, we will use the following equation of motion:
Vyi = Voy + (-g)t
where Voy is the vertical component of the initial velocity, g is the acceleration due to gravity (approximately -9.8 m/s^2), and t is the time taken.

68.70 m/s = 0 m/s + (-9.8 m/s^2)t
t = 68.70 m/s / 9.8 m/s^2
t ≈ 7.04 s

Step 3: Find the maximum height reached by the rocket:
The maximum height can be calculated using the formula:
Hmax = Voy * t + (1/2) * g * t^2
where Hmax is the maximum height reached, Voy is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time taken.

Hmax = 68.70 m/s * 7.04 s + (1/2) * (-9.8 m/s^2) * (7.04 s)^2
Hmax = 482.328 m + (-242.616 m)
Hmax ≈ 239.71 m

Step 4: Find the time it takes for the rocket to travel horizontally to the wall:
The time taken to reach the wall can be calculated using the formula:
t = d / Vx
where d is the horizontal distance to the wall (25.0 m) and Vx is the horizontal component of the initial velocity.

Vx = V * cos(θ)
Vx = 91.0 m/s * cos(49.0°)
Vx = 91.0 m/s * 0.6561
Vx ≈ 59.78 m/s

t = 25.0 m / 59.78 m/s
t ≈ 0.42 s

Step 5: Find the vertical distance traveled by the rocket during the time it takes to reach the wall:
During the time taken to reach the wall, the rocket descends from the maximum height to the top of the wall. The vertical distance traveled can be calculated using the formula:
d = Voy + (-g) * t
where d is the vertical distance traveled, Voy is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time taken.

d = 68.70 m/s * 0.42 s + (-9.8 m/s^2) * (0.42 s)^2
d = 28.83 m + (-1.72 m)
d ≈ 27.11 m

Step 6: Find the clearance of the rocket over the top of the wall:
The clearance can be calculated by subtracting the height of the wall (21.4 m) from the vertical distance traveled by the rocket.

Clearance = d - height of the wall
Clearance = 27.11 m - 21.4 m
Clearance ≈ 5.71 m

Therefore, the rocket clears the top of the wall by approximately 5.71 meters.