A tennis player standing 13.9 m from the net hits the ball at 3.40° above the horizontal. To clear the net, the ball must rise at least 0.449 m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racquet?
Yf^2 = Yo + 2g*h = 0.
Yo = -2g*h = -2*(-9.8)*0.449 = 8.8 m/s.
Yo = Vo*sin3.40 = 8.8 m/s.
Vo = 8.8/sin3.4 = 148 m/s.
To find the initial speed of the ball when it left the racquet, we can use the principles of projectile motion. Let's break down the problem into its different components:
1. Vertical Motion:
The ball must rise at least 0.449 m above the initial height of the net. We can use the equation for vertical displacement:
Δy = v₀y * t + (1/2) * a * t²
Where:
Δy = vertical displacement (0.449 m)
v₀y = initial vertical velocity of the ball (unknown)
t = time of flight (unknown)
a = acceleration due to gravity (-9.8 m/s²)
Since the ball reaches its maximum height at the apex, the final vertical velocity (vfy) will be zero. Therefore, the equation becomes:
0 = v₀y + (-9.8) * t
This equation can be rearranged to solve for t:
t = v₀y / 9.8
2. Horizontal Motion:
The horizontal motion of the ball remains unaffected by gravity. We can use the equation for horizontal displacement:
Δx = v₀x * t
Where:
Δx = horizontal displacement (13.9 m)
v₀x = initial horizontal velocity of the ball (unknown)
3. Projected Angle:
We know that the ball was hit at an angle of 3.40° above the horizontal. This angle can be used to find the vertical and horizontal components of the initial velocity:
v₀y = v₀ * sinθ
v₀x = v₀ * cosθ
Where:
v₀ = initial velocity of the ball (unknown)
θ = angle of projection (3.40°)
Combining all the equations, we can solve for v₀:
Δx = (v₀ * cosθ) * (v₀y / 9.8)
Substituting the values for Δx and θ:
13.9 = (v₀ * cos(3.40°)) * (v₀ * sin(3.40°) / 9.8)
Now, we can solve this equation to find the initial speed (v₀) of the ball when it left the racquet.