Physics Notes: (filled with academic musings)
Time of flight (t):
From the horizontal motion equation, we have:
t = Δx / v_x
Initial vertical velocity (v_iy):
Substitute the expression for t into the vertical motion equation:
Δy = v_iy * (Δx / v_x) + (1/2) * g * (Δx / v_x)²
Horizontal and vertical velocity components:
v_iy = v_x * tan(40°)
Substitute and solve for v_x:
Substitute v_iy in the equation for Δy:
Δy = (v_x * tan(40°)) * (Δx / v_x) + (1/2) * g * (Δx² / v_x²)
Δy = Δx * tan(40°) + (g * Δx²) / (2 * v_x²)
Now, rearrange to solve for v_x:
(g * Δx²) / (2 * v_x²) = Δy - Δx * tan(40°)
v_x² = (g * Δx²) / (2 * (Δy - Δx * tan(40°)))
v_x = sqrt((g * Δx²) / (2 * (Δy - Δx * tan(40°))))
Angle of rebound: The ball rebounds at certain degrees such as 40° from the horizontal.
Vertical velocity at bounce: Let v be the rebound speed. The vertical velocity component immediately after the bounce is v_y = v * sin(40°).
Time to reach apex: The time it takes for the ball to reach the highest point of its trajectory is when the vertical velocity becomes zero. Using the equation v_f = v_i + at: 0 = v * sin(40°) - g * t_up t_up = (v * sin(40°)) / g
Total time of flight: The total time between bounces is twice the time to reach the apex (due to symmetry): t = 2 * t_up = (2 * v * sin(40°)) / g
Horizontal distance: The horizontal distance is given by: Δx = v_x * t = (v * cos(40°)) * ((2 * v * sin(40°)) / g) Δx = (v² * sin(80°)) / g (using the trigonometric identity 2sinθcosθ = sin2θ)
Solve for v: v² = (Δx * g) / sin(80°) v = sqrt((Δx * g) / sin(80°))