To solve this problem, we can use the equations of motion for projectile motion. Let's break down each part step-by-step:
Step 1: Find the time of flight
The ball is thrown up and lands back down, so the total time of flight is twice the time it takes for the ball to reach its peak. Given that the time it takes to reach the roof is 4 seconds, the time of flight is 2 * 4 = 8 seconds.
Step 2: Find the initial vertical velocity
Since the ball starts and ends at the same height, and the path is at an angle, we can assume there is no change in the vertical direction. Therefore, the initial vertical velocity is 0 m/s.
Step 3: Find the vertical distance traveled
Using the equation for vertical displacement, we have:
d = Vâ‚€y * t + (1/2) * g * t^2
where d is the vertical distance (20m), Vâ‚€y is the initial vertical velocity, t is the time of flight (8s), and g is the acceleration due to gravity.
Since Vâ‚€y = 0, the equation simplifies to:
20 = (1/2) * g * t^2
Solving for g:
2 * 20 = g * 8^2
g = 5 m/s²
Step 4: Find the horizontal distance
The horizontal distance is given by:
d = Vâ‚€x * t
where d is the horizontal distance, Vâ‚€x is the initial horizontal velocity, and t is the time of flight.
We can use the fact that the horizontal and vertical displacements are related:
d = V₀x * t = V₀ * cosθ * t
where V₀ is the initial velocity and θ is the angle with respect to the horizontal.
Substituting the values:
20 = V₀ * cos(60°) * 8
Solving for Vâ‚€:
V₀ = 20 / (8 * cos(60°))
Step 5: Calculate the magnitude of the initial velocity
The magnitude of the initial velocity is given by the formula:
|Vâ‚€| = sqrt(Vâ‚€x^2 + Vâ‚€y^2)
where Vâ‚€ is the initial velocity, Vâ‚€x is the initial horizontal velocity, and Vâ‚€y is the initial vertical velocity.
Since Vâ‚€y = 0, the equation simplifies to:
|Vâ‚€| = |Vâ‚€x|
Substituting the values:
|Vâ‚€| = sqrt(Vâ‚€x^2)
Step 6: Calculate the angle of the initial velocity
The angle θ₀ of the initial velocity is given by the formula:
θ₀ = arctan(V₀y / V₀x)
Since Vâ‚€y = 0, the angle simplifies to:
θ₀ = arctan(0 / V₀x) = arctan(0) = 0
a) The horizontal distance traveled by the ball is given by d = V₀ * cosθ * t. Substituting the values, we get d = (20 / (8 * cos(60°))) * 8.
b) The magnitude of the ball's initial velocity is given by |V₀| = sqrt(V₀x^2). Substituting the values, we get |V₀| = sqrt((20 / (8 * cos(60°)))^2).
c) The angle of the ball's initial velocity with respect to the horizontal is given by θ₀ = arctan(0) = 0 degrees.