A cannonball is shot (from ground level) with an initial horizontal velocity of 38.0 m/s and an initial vertical velocity of 26.0 m/s. The initial speed of the cannonball 46.04 m/s The initial angle θ of the cannonball with respect to the ground 34.38°

What is the maximum height the cannonball goes above the ground? (m)
How far from where it was shot will the cannonball land? (m)
What is the speed of the cannonball 2.7 seconds after it was shot? (m/s)
How high above the ground is the cannonball 2.7 seconds after it is shot? (m)

34.2 meters

To solve these questions, we can use the equations of projectile motion. The key equations we will use are:

1. Vertical displacement (height):
d = v0y * t + 0.5 * a * t^2

2. Horizontal displacement (range):
R = v0x * t

3. Velocity components (horizontal and vertical):
v0x = v0 * cos(θ)
v0y = v0 * sin(θ)

4. Final velocity components:
vx = v0x
vy = v0y + a * t

Now let's solve each question step by step.

1. Maximum height the cannonball goes above the ground:
The vertical displacement at the maximum height will be zero (ball reaches its peak). We can use the vertical displacement equation with v0y as the initial vertical velocity and a as the acceleration due to gravity (-9.8 m/s^2).
Setting d = 0, we have:
0 = v0y * t + 0.5 * (-9.8) * t^2
Simplifying the equation, we get:
0.5 * (-9.8) * t^2 = v0y * t
Rearranging the equation, we get:
t = (2 * v0y) / 9.8
Substituting the given values, we get:
t = (2 * 26.0) / 9.8 ≈ 5.31 seconds
Now that we have the time of flight, we can substitute it back into the vertical displacement equation to find the maximum height:
d = v0y * t + 0.5 * (-9.8) * t^2
d = 26.0 * 5.31 + 0.5 * (-9.8) * (5.31)^2
d ≈ 69.18 meters
Therefore, the maximum height the cannonball goes above the ground is approximately 69.18 meters.

2. Distance from where the cannonball was shot to where it will land:
The horizontal displacement is given by the range equation. We can calculate it using the initial horizontal velocity (v0x) and the time of flight (t) which we found in the previous question.
R = v0x * t
R = 38.0 * 5.31
R ≈ 201.78 meters
Therefore, the cannonball will land approximately 201.78 meters away from where it was shot.

3. Speed of the cannonball 2.7 seconds after it was shot:
The speed of the cannonball is the magnitude of its velocity vector at a given time. We can calculate it using the horizontal and vertical components of the velocity. At any given time, the vertical component of velocity remains constant, while the horizontal component remains unchanged (no horizontal acceleration). The magnitude of the velocity can be calculated using the Pythagorean theorem.
vx = v0x
vy = v0y + a * t
v = sqrt(vx^2 + vy^2)
v = sqrt(v0x^2 + (v0y + a * t)^2)
v = sqrt((v0 * cos(θ))^2 + (v0 * sin(θ) + a * t)^2)
v = sqrt((46.04 * cos(34.38))^2 + (46.04 * sin(34.38) - 9.8 * 2.7)^2)
v ≈ 37.61 meters per second
Therefore, the speed of the cannonball 2.7 seconds after it was shot is approximately 37.61 m/s.

4. Height above the ground 2.7 seconds after it is shot:
Similar to the first question, we can use the vertical displacement equation to find the height. We need to substitute the initial vertical velocity (v0y), time (t), and acceleration due to gravity (a).
d = v0y * t + 0.5 * a * t^2
d = 26.0 * 2.7 + 0.5 * (-9.8) * (2.7)^2
d ≈ 11.14 meters
Therefore, the cannonball is approximately 11.14 meters above the ground 2.7 seconds after it is shot.