When John Wilkins hits a baseball at an angle of 35° with an initial velocity of 110 mph, did he hit a home run if the fence is 403 ft away?

In evaluating whether John Wilkins hit a home run when he hit the baseball at a 35° angle with an initial velocity of 110 mph, we can calculate the horizontal distance the baseball would travel. This distance can be determined using the principles of projectile motion.

First, we need to convert the speed of the ball into feet per second since the distance to the fence is given in feet. One mile per hour is equal to approximately 1.46667 feet per second, so 110 mph converts to about 161.0 feet per second.

Next, calculate the horizontal and vertical components of the velocity. The horizontal component can be found using the cosine of the launch angle, while the vertical component can be found using the sine of the launch angle:

• Horizontal velocity ( V_{x} = V \cdot \cos(\theta) )

• Vertical velocity ( V_{y} = V \cdot \sin(\theta) )

For an angle of 35°, the calculations yield:

• ( V_{x} \approx 161.0 \cdot \cos(35°) )

• ( V_{y} \approx 161.0 \cdot \sin(35°) )

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