A tennis ball is served 2.00 degrees above the horizontal at a height of 2.40 meters, 12.0 meters from
a net that is 0.900 meters high.
(a) If the tennis ball is to clear the net by at least 0.200 meters, what is its minimum initial velocity?
(b) If the tennis ball clears the net by 0.200 meters, where will it land?
To solve this problem, we can use kinematic equations and the principles of projectile motion. Here's how you can find the answers to both parts of the question:
(a) To find the minimum initial velocity for the tennis ball to clear the net by at least 0.200 meters, we need to consider the vertical motion of the ball. The key equation we can use is the kinematic equation for vertical displacement:
Δy = v₀y * t + (1/2) * g * t²
where:
Δy = vertical displacement (0.200 m in this case)
v₀y = initial vertical velocity (what we're trying to find)
t = time of flight (which can be determined from the horizontal motion)
g = acceleration due to gravity (approximately 9.8 m/s²)
First, let's find the time of flight. We can consider the horizontal motion of the ball, where the acceleration is zero. The horizontal equation is:
x = v₀x * t
where:
x = horizontal distance (12.0 m in this case)
v₀x = initial horizontal velocity
t = time of flight
To find the horizontal velocity, we can use trigonometry. The horizontal velocity can be calculated using the initial velocity and the launch angle:
v₀x = v₀ * cos(θ)
where:
v₀ = initial velocity (what we're trying to find)
θ = launch angle (2.00 degrees)
Now we have enough information to find the minimum initial velocity. Let's go step by step:
Step 1: Calculate the time of flight (t) from the horizontal motion equation:
12.0 m = v₀x * t
Solve for t: t = 12.0 m / v₀x
Step 2: Calculate the initial horizontal velocity (v₀x) using trigonometry:
v₀x = v₀ * cos(θ)
Step 3: Substitute the expression for v₀x obtained in step 2 into the equation in step 1:
t = 12.0 m / (v₀ * cos(θ))
Step 4: Substitute the expression for t obtained in step 3 into the vertical motion equation:
Δy = v₀y * (12.0 m / (v₀ * cos(θ))) + (1/2) * g * (12.0 m / (v₀ * cos(θ)))²
Step 5: Rearrange the equation and solve for v₀ (minimum initial velocity):
Multiply through by (v₀ * cos(θ))², then solve for v₀:
(v₀ * cos(θ))² * Δy = v₀y * (12.0 m)² + (1/2) * g * (12.0 m)²
v₀² = (v₀y * (12.0 m)² + (1/2) * g * (12.0 m)²) / (cos(θ)² * Δy)
v₀ = √((v₀y * (12.0 m)² + (1/2) * g * (12.0 m)²) / (cos(θ)² * Δy))
Plug in the given values for Δy, g, θ, and solve for v₀ to find the minimum initial velocity.
(b) To find where the tennis ball will land if it clears the net by 0.200 meters, we need to consider the horizontal motion of the ball. Since there is no horizontal acceleration, the horizontal distance traveled will be the same as the horizontal distance from the launch point to the net.
Using the same horizontal motion equation as before:
x = v₀x * t
We can substitute the values we already know:
x = v₀ * cos(θ) * t
To find the total time of flight, we can use the known vertical displacement and initial velocity:
Δy = v₀y * t + (1/2) * g * t²
Rearrange the equation and solve for t:
t = (√((2 * Δy) / g) - v₀y) / g
Now, substitute the expression for t into the equation for x:
x = v₀ * cos(θ) * [(√((2 * Δy) / g) - v₀y) / g]
Plug in the known values for Δy, g, θ, v₀, and v₀y to find the horizontal distance (x) where the ball will land.