When Babe Ruth hit a homer over the 10 m -high right-field fence 80 m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.2 m above the ground and its path initially made a 39 ∘ angle with the ground.
To find the minimum speed of the ball when it left the bat, we can use the principles of projectile motion. We can break down the motion of the ball into its horizontal and vertical components.
Step 1: Find the time of flight (horizontal motion)
Since we are given the horizontal distance traveled by the ball, we can use the formula:
Horizontal distance = Horizontal velocity x Time of flight
In this case, the horizontal distance is 80 m. We need to find the time of flight.
The horizontal velocity can be calculated by multiplying the initial velocity of the ball by the cosine of the launch angle.
Horizontal velocity = initial velocity x cos(angle)
Step 2: Find the vertical motion
We are given that the ball cleared a 10 m-high fence. The vertical distance covered by the ball can be calculated using the following formula:
Vertical distance = Vertical velocity x Time of flight + (1/2) x Acceleration x (Time of flight)^2
In this case, the vertical distance is 10 m. We need to find the vertical velocity and time of flight.
The vertical velocity can be calculated by multiplying the initial velocity of the ball by the sine of the launch angle.
Vertical velocity = initial velocity x sin(angle)
Step 3: Combine the equations
Since the time of flight is the same for both horizontal and vertical motion, we can equate the two equations:
Horizontal distance = Vertical velocity x Time of flight
Now we can solve for the time of flight.
Step 4: Find the minimum speed
The minimum speed is the magnitude of the initial velocity of the ball, which can be calculated using the formula:
Initial velocity = √(Horizontal velocity^2 + Vertical velocity^2)
Now, let's perform the calculations:
Given:
Horizontal distance = 80 m
Vertical distance = 10 m
Launch angle = 39°
Step 1:
Horizontal velocity = initial velocity x cos(angle)
Vertical velocity = initial velocity x sin(angle)
Step 2:
Vertical distance = Vertical velocity x Time of flight + (1/2) x Acceleration x (Time of flight)^2
Step 3:
Horizontal distance = Vertical velocity x Time of flight
Step 4:
Initial velocity = √(Horizontal velocity^2 + Vertical velocity^2)
After completing these calculations, you should find the minimum speed of the ball when it left the bat.
To calculate the minimum speed of the ball when it left the bat, we can use the principle of conservation of energy.
1. First, let's find the initial vertical velocity component of the ball. The ball was hit 1.2 m above the ground, so we can calculate the initial vertical velocity, Vy.
Given:
Initial vertical position (y0) = 1.2 m
Gravitational acceleration (g) = 9.8 m/s^2
The formula to calculate the initial vertical velocity component is:
Vy = sqrt(2 * g * (y - y0))
Vy = sqrt(2 * 9.8 * (10 - 1.2))
Vy = sqrt(2 * 9.8 * 8.8)
Vy ≈ sqrt(172.16)
Vy ≈ 13.11 m/s
2. Next, let's find the initial horizontal velocity component of the ball. The path of the ball made a 39° angle with the ground, so we can calculate the initial horizontal velocity, Vx.
Given:
Launch angle (θ) = 39°
The formula to calculate the initial horizontal velocity component is:
Vx = V * cos(θ)
Vx = V * cos(39°)
3. Now, we can find the total initial velocity (V) of the ball using the Pythagorean theorem, which relates the vertical and horizontal components of the velocity.
Given:
Total initial velocity (V) = sqrt(Vx^2 + Vy^2)
Using the values calculated above:
V = sqrt((Vx)^2 + (Vy)^2)
V = sqrt((V * cos(39°))^2 + (13.11 m/s)^2)
We have an equation in terms of V, let's solve for V:
V = sqrt((V * cos(39°))^2 + (13.11 m/s)^2)
Simplifying the equation:
V = sqrt(V^2 * cos^2(39°) + (13.11 m/s)^2)
V^2 = V^2 * cos^2(39°) + (13.11 m/s)^2
0 = V^2 * cos^2(39°) - V^2 + (13.11 m/s)^2
0 = V^2 * (cos^2(39°) - 1) + (13.11 m/s)^2
V^2 = ((13.11 m/s)^2) / (1 - cos^2(39°))
V = sqrt(((13.11 m/s)^2) / (1 - cos^2(39°)))
Using a calculator:
V ≈ sqrt(171.09245 / 0.248150)
V ≈ sqrt(690.45995)
V ≈ 26.27 m/s
Therefore, the minimum speed of the ball when it left the bat was approximately 26.27 m/s.