To solve this problem, we can break it down into different components and use the equations of motion to find the answers.
(a) Find the speed at which the ball was launched:
To find the speed, we can use the horizontal component of the initial velocity since the vertical component is not given. We can use the equation:
v = d / t
where v is the velocity, d is the horizontal distance, and t is the time taken.
Given: d = 24.0 m and t = 2.20 s
Plugging in these values, we get:
v = 24.0 m / 2.20 s
v ≈ 10.91 m/s
Therefore, the speed at which the ball was launched is approximately 10.91 m/s.
(b) Find the vertical distance by which the ball clears the wall:
To determine the vertical distance the ball clears the wall, we need to find the vertical component of the projectile's velocity. We can use the equation:
y = y0 + v0y * t + (1/2) * a * t^2
where y is the vertical distance, y0 is the initial vertical position, v0y is the initial vertical velocity, t is the time taken, and a is the acceleration (in this case, due to gravity).
Given: y0 = 5.8 m, t = 2.20 s, and a = -9.8 m/s^2 (acceleration due to gravity)
To find v0y, the initial vertical velocity, we can use the equation:
v0y = v * sin(θ)
where v is the initial speed and θ is the launch angle.
v = 10.91 m/s (as found in part (a))
θ = 53.0°
Plugging in these values, we get:
v0y = 10.91 m/s * sin(53.0°)
v0y ≈ 8.71 m/s
Now we can find the vertical distance by plugging the values into the equation:
y = 5.8 m + (8.71 m/s) * (2.20 s) + (1/2) * (-9.8 m/s^2) * (2.20 s)^2
y ≈ 5.8 m + 19.16 m - 9.71 m
y ≈ 15.25 m
Therefore, the ball clears the wall by approximately 15.25 meters vertically.
(c) Find the horizontal distance from the wall to the point on the roof where the ball lands:
To find the horizontal distance, we can use the equation:
x = v0x * t
where x is the horizontal distance, v0x is the initial horizontal velocity, and t is the time taken.
To find v0x, the initial horizontal velocity, we can use the equation:
v0x = v * cos(θ)
Given: v = 10.91 m/s (as found in part (a)), θ = 53.0°, and t = 2.20 s
Plugging in these values, we get:
v0x = 10.91 m/s * cos(53.0°)
v0x ≈ 6.21 m/s
Now we can find the horizontal distance by plugging the values into the equation:
x = (6.21 m/s) * (2.20 s)
x ≈ 13.65 m
Therefore, the horizontal distance from the wall to the point on the roof where the ball lands is approximately 13.65 meters.