a. To solve for the time it takes the ball to reach the back wall, we need to use the kinematic equations of motion for projectile motion. We can start by breaking the initial velocity into its horizontal and vertical components:
vx = v0 cosθ = 100 ft/s cos(30°) ≈ 86.6 ft/s
vy = v0 sinθ = 100 ft/s sin(30°) ≈ 50 ft/s
where v0 is the initial velocity and θ is the angle with the horizontal.
The horizontal motion is uniform and does not affect the time it takes for the ball to reach the back wall, so we can focus on the vertical motion:
y = y0 + v0t sinθ - 1/2gt^2
where y is the vertical displacement, y0 is the initial vertical position, t is the time, g is the acceleration due to gravity (approximately 32.2 ft/s^2).
We know that the initial vertical position is 3 feet, and we want to find the time it takes for the ball to reach a vertical position of 9 feet (the height of the back wall). Therefore, we can plug in these values and solve for t:
9 ft = 3 ft + (100 ft/s sin(30°))t - 1/2 (32.2 ft/s^2)t^2
0 = -1/2 (32.2 ft/s^2)t^2 + (50 ft/s)t + 6 ft
Using the quadratic formula:
t ≈ 3.27 seconds
Therefore, it takes the ball approximately 3.27 seconds to reach the back wall (when no one interferes with it).
b. To determine if the ball goes over the back wall, we need to find the horizontal distance the ball travels during the time it takes to reach the height of the back wall. We can use the horizontal velocity we found earlier and the time we just calculated:
x = vxt
x = (86.6 ft/s)(3.27 s)
x ≈ 283.1 feet
Since the back wall is 200 feet away, the ball does not go over the back wall. It lands approximately 83.1 feet in front of the back wall.