Babe Didrikson holds the world record for the longest baseball throw (296 ft) by a woman. For the following questions, assume that a ball is thrown at an angle of 45.0° above the horizontal, that it travels a horizontal distance of 280 ft, and is caught at the same level from which it was thrown.
What is the ball's initial speed? How long is the ball in the air?
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To find the ball's initial speed, we can use the horizontal distance traveled and the launch angle.
Given:
Horizontal distance (range) = 280 ft
Launch angle = 45.0°
The horizontal distance traveled by the ball is equal to its initial horizontal velocity multiplied by the time of flight. Since there is no horizontal acceleration, the vertical motion can be disregarded.
Horizontal distance = (initial velocity * cos launch angle) * time
Since the launch angle is 45.0°, the cosine of 45.0° is 0.7071.
280 ft = (initial velocity * 0.7071) * time
We can solve for the initial velocity by rearranging the equation:
initial velocity * time = (280 ft / 0.7071)
initial velocity = (280 ft / 0.7071) / time
Now, to find the time of flight, we can use the vertical component of the projectile motion. At the highest point of the throw, the vertical velocity will be zero.
Using the equation:
Vertical velocity = initial vertical velocity - (gravitational acceleration * time)
Since the gravitational acceleration is 32.2 ft/s^2 and the launch angle is 45.0°, the initial vertical velocity is equal to the initial velocity * sin launch angle. Similarly, the time of flight is equal to 2 * (initial velocity * sin launch angle) / gravitational acceleration.
Let's calculate the values step by step:
1. Calculate the initial velocity:
initial velocity = (280 ft / 0.7071) / time
2. Calculate the time of flight:
time of flight = 2 * initial velocity * sin launch angle / gravitational acceleration
Note: I'll use imperial units in the calculations, but the values can be converted to any other unit if desired.
Now, if you provide the time (in seconds) or require further assistance, I can provide more accurate calculations.
To determine the ball's initial speed, we can use the range formula for projectile motion. The range (R) of a projectile is given by the equation:
R = (v^2 * sin(2θ)) / g
where:
- v is the initial velocity (speed)
- θ is the launch angle
- g is the acceleration due to gravity (approximately 32.2 ft/s^2)
We are given that the launch angle is 45.0° and the horizontal distance traveled (range) is 280 ft. Plugging these values into the formula, we can solve for v.
280 ft = (v^2 * sin(90°)) / (32.2 ft/s^2)
sin(90°) = 1 (since sin(90°) = 1)
280 ft = (v^2 * 1) / (32.2 ft/s^2)
280 ft * 32.2 ft/s^2 = v^2
v^2 = 9026 ft^2/s^2
v ≈ √9026 ft/s
v ≈ 95.02 ft/s
Therefore, the ball's initial speed is approximately 95.02 ft/s.
To determine how long the ball is in the air, we can use the time of flight formula:
t = (2 * v * sin(θ)) / g
Plugging in the known values:
t = (2 * 95.02 ft/s * sin(45.0°)) / 32.2 ft/s^2
t ≈ (2 * 95.02 ft/s * 0.7071) / 32.2 ft/s^2
t ≈ (134.27 ft/s) / 32.2 ft/s^2
t ≈ 4.17 s (rounded to two decimal places)
Therefore, the ball is in the air for approximately 4.17 seconds.