While warming up in her tryout as a pitcher for the Montwood softball team, Anabel Parga throws a softball horizontally with a speed of 22m/s.
a. If home plate is 19m away, how far should the softball fall from its initial straight-line path on its way to home plate?
b. The softball is thrown from a height of 2.1m. What is the displacement from home plate of the softball when it strikes the ground?
To solve these questions, we'll use the equations of motion for projectile motion. In projectile motion, an object moves in a curved path under the influence of gravity.
a. To find how far the softball falls from its initial straight-line path, we need to determine the vertical displacement.
The equation for vertical displacement in projectile motion is given by:
Δy = v₀y * t + (1/2) * a * t²
Here, Δy is the vertical displacement, v₀y is the initial vertical velocity (which is zero as the softball is thrown horizontally), a is the acceleration due to gravity (-9.8 m/s²), and t is the time it takes for the softball to reach home plate.
To find the time, we use the equation:
Δx = v₀x * t
where Δx is the horizontal displacement (19m in this case) and v₀x is the initial horizontal velocity (22m/s in this case).
Rearranging the equation to solve for time, we have:
t = Δx / v₀x
Substituting the given values:
t = 19m / 22m/s
t ≈ 0.864 seconds
Now, let's calculate the vertical displacement:
Δy = 0 * 0.864 + (1/2) * (-9.8 m/s²) * (0.864)²
Δy ≈ -3.94 meters
Therefore, the softball falls approximately 3.94 meters from its initial straight-line path on its way to home plate.
b. To find the displacement from home plate when the softball strikes the ground, we need to calculate the time it takes for the softball to hit the ground.
The equation for time of flight (the time it takes to reach the ground) is:
t = √(2 * Δy / a)
Using the vertical displacement Δy as -2.1m (since it's a downward displacement),
t = √(2 * (-2.1m) / -9.8 m/s²)
t ≈ 0.62 seconds
Now, let's calculate the horizontal displacement:
Δx = v₀x * t
Δx = 22m/s * 0.62s
Δx ≈ 13.64 meters
Therefore, the displacement of the softball from home plate when it strikes the ground is approximately 13.64 meters.
To find the answers to these questions, we can use the equations of motion in the horizontal and vertical directions. Let's solve them step-by-step:
a. To find how far the softball should fall from its initial straight-line path on its way to home plate, we need to find the time it takes for the softball to reach home plate first.
Step 1: Identify the given variables:
- Initial horizontal speed (Vi) = 22 m/s
- Horizontal distance (d) = 19 m
- Acceleration in the horizontal direction (a) = 0 (as there is no acceleration in the horizontal direction)
Step 2: Use the equation of motion for horizontal distance:
d = Vi * t
19m = 22m/s * t
Step 3: Solve for time (t):
t = 19m / 22m/s
t ≈ 0.864 s
Now that we have the time it takes for the softball to reach home plate (t), we can find how far it falls from its initial straight-line path.
Step 4: Use the equation of motion for vertical distance:
Vertical distance (dy) = (1/2) * g * t^2
(assuming the acceleration due to gravity, g, is approximately 9.8 m/s^2)
dy = (1/2) * 9.8m/s^2 * (0.864s)^2
Step 5: Solve for vertical distance (dy):
dy ≈ 3.68 m
So, the softball should fall approximately 3.68 meters from its initial straight-line path on its way to home plate.
b. To find the displacement from home plate of the softball when it strikes the ground, we need to find the time it takes for the softball to reach the ground.
Step 1: Identify the given variables:
- Initial vertical height (h) = 2.1 m
- Acceleration in the vertical direction (a) = -9.8 m/s^2 (negative because gravity acts downward)
Step 2: Use the equation of motion for vertical displacement:
h = Vi * t + (1/2) * a * t^2
2.1m = 0 * t + (1/2) * (-9.8m/s^2) * t^2
Step 3: Solve for time (t):
2.1m = -4.9m/s^2 * t^2
t^2 ≈ 0.42857
t ≈ √0.42857
t ≈ 0.654 s
Now that we have the time it takes for the softball to reach the ground (t), we can find the displacement from home plate.
Step 4: Use the equation of motion for horizontal distance:
Horizontal distance (dx) = Vi * t
dx = 22m/s * 0.654s
Step 5: Solve for horizontal distance (dx):
dx ≈ 14.508 m
Since we assumed the horizontal distance (dx) is positive, the displacement from home plate is approximately 14.508 meters.
Therefore, the softball falls approximately 3.68 meters from its initial straight-line path, and the displacement from home plate is approximately 14.508 meters when it strikes the ground.