Respuesta :
To solve this problem, we can use the kinematic equations of motion. The horizontal and vertical motion of the ball can be treated as separate motions.
First, let's find the time it takes for the ball to reach the wall. The horizontal distance the ball travels can be found using the equation:
x = v0 * cos(θ) * t
where x = 130.7m, v0 is the initial speed, θ = 36.1°, and t is the time it takes to reach the wall.
Solving for t:
130.7m = v0 * cos(36.1°) * t
t = 130.7m / (v0 * cos(36.1°))
Now, let's find the vertical distance the ball travels. The vertical distance can be found using the equation:
y = v0 * sin(θ) * t - 0.5 * g * t^2
where g = 9.81m/s^2 is the acceleration due to gravity.
The ball's vertical motion can be separated into two parts: reaching the maximum height and then falling to the ground.
The maximum height can be found using the equation:
v_f^2 = v_i^2 - 2 * g * Δy
where v_f = 0 (at the maximum height), v_i is the initial vertical velocity, g = 9.81m/s^2, and Δy is the vertical displacement.
v_i^2 = 2 * g * Δy
v_i = sqrt(2 * g * Δy)
Now, let's find the time it takes for the ball to reach the maximum height. The time can be found using the equation:
v_f = v_i - g * t
0 = v_i - g * t
t = v_i / g
Substitute v_i and solve:
t = sqrt(2 * Δy / g) / g
The total time of flight will be twice this time since the ball takes the same amount of time to rise to its maximum height as it does to fall back down.
Now we have everything we need to solve for the initial speed v0.
t = 130.7m / (v0 * cos(36.1°))
v_i = sqrt(2 * g * Δy)
t = 2 * sqrt((2 * Δy) / g)
Plugging in the values for Δy = 8.0m and g = 9.81m/s^2, we can solve for v0.
t = 130.7m / (v0 * cos(36.1°)) = 2 * sqrt((2 * 8.0m) / 9.81m/s^2)
Solving for v0:
v0 = 31.3 m/s
So, the initial speed of the ball is 31.3 m/s.
