To solve this problem, we can analyze the horizontal and vertical components of the ball's motion separately.
First, let's determine the time it takes for the ball to reach the wall. The horizontal component of the ball's velocity remains constant throughout its motion. The horizontal distance traveled is 10 m, and the initial horizontal velocity is given by:
Vx = V * cos(θ)
where V is the initial velocity (20.0 m/s) and θ is the angle of projection (65 degrees). Substituting the values:
Vx = 20.0 m/s * cos(65 degrees) Vx = 20.0 m/s * 0.42262 Vx ≈ 8.45 m/s
The time taken to cover the horizontal distance can be calculated using the equation:
time = distance / velocity
time = 10 m / 8.45 m/s time ≈ 1.18 s
Now let's determine the height at which the ball will strike the wall. The vertical motion can be analyzed using the equation:
y = y0 + Vy0 * t - (1/2) * g * t^2
where y is the vertical position, y0 is the initial vertical position (1.8 m), Vy0 is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The initial vertical velocity can be calculated using:
Vy0 = V * sin(θ)
Vy0 = 20.0 m/s * sin(65 degrees) Vy0 = 20.0 m/s * 0.90631 Vy0 ≈ 18.13 m/s
Substituting the values into the equation:
y = 1.8 m + 18.13 m/s * 1.18 s - (1/2) * 9.8 m/s^2 * (1.18 s)^2 y ≈ 1.8 m + 21.4 m - 6.19 m y ≈ 16.01 m
Therefore, the ball will strike the 20 m tall wall at a height of approximately 16.01 meters above the ground.