In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 26m/s at an angle of 35^\circ above the horizontal.How high is the ball when it hits the wall?

I wonder how far away is the wall?

3.9

To determine the height of the ball when it hits the wall, we can use the projectile motion equations. Here's how:

Step 1: Resolve the initial velocity into horizontal and vertical components.
The horizontal component (Vx) can be found using the formula: Vx = V * cos(θ).
Vx = 26 m/s * cos(35°)
Vx = 26 m/s * 0.8192
Vx ≈ 21.27 m/s

The vertical component (Vy) can be found using the formula: Vy = V * sin(θ).
Vy = 26 m/s * sin(35°)
Vy = 26 m/s * 0.574
Vy ≈ 14.92 m/s

Step 2: Determine the time of flight.
The time the ball takes to reach the wall can be calculated using the vertical component of the velocity. Since there is no vertical acceleration (assuming no air resistance or other external forces), we can use the formula: t = 2 * Vy / g.
g is the acceleration due to gravity, which is approximately 9.8 m/s².
t = 2 * 14.92 m/s / 9.8 m/s²
t ≈ 3.04 s

Step 3: Calculate the height.
The vertical displacement (Δy) of the ball when it hits the wall can be calculated using the formula: Δy = Vy * t - 0.5 * g * t².
Δy = 14.92 m/s * 3.04 s - 0.5 * 9.8 m/s² * (3.04 s)²
Δy ≈ 45.37 m

Therefore, the ball is approximately 45.37 meters above the ground when it hits the wall.