According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy “Dizzy” Carlyle in a minor league game. The ball traveled 188 {\rm m} (618 {\rm ft}) before landing on the ground outside the ballpark
part a
Assuming the ball's initial velocity was 44^\circ above the horizontal and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 {\rm m} (3.0 {\rm ft}) above ground level? Assume that the ground was perfectly flat.
Part B
How far would the ball be above a fence 3.0 {\rm m} (10 {\rm ft}) high if the fence was 116 {\rm m} (380 {\rm ft}) from home plate?
Part A:
To find the initial speed of the ball, we can use the range formula for projectile motion:
R = (V^2 * sin(2θ)) / g
where:
R is the range (distance traveled by the ball),
V is the initial velocity of the ball,
θ is the launch angle (44 degrees in this case),
and g is the acceleration due to gravity (assumed to be 9.8 m/s^2).
Given:
R = 188 m (618 ft),
θ = 44 degrees,
h = 0.9 m (3.0 ft),
and g = 9.8 m/s^2,
we can rearrange the formula to solve for V:
V = sqrt((R * g) / sin(2θ))
Plugging in the known values, we get:
V = sqrt((188 * 9.8) / sin(2 * 44))
Calculating this, we find:
V ≈ 41.7 m/s
Therefore, the initial speed of the ball needed to produce such a home run is approximately 41.7 m/s.
Part B:
To find how far above a fence the ball would be at a given distance, we can use the height formula for projectile motion:
h = h0 + (V^2 * sin^2(θ)) / (2g)
where:
h is the vertical distance above the ground (or fence),
h0 is the initial height (0.9 m in this case),
V is the initial velocity of the ball,
θ is the launch angle (44 degrees in this case),
and g is the acceleration due to gravity (assumed to be 9.8 m/s^2).
Given:
h0 = 0.9 m (3.0 ft),
V = 41.7 m/s,
θ = 44 degrees,
d = 116 m (380 ft),
and g = 9.8 m/s^2,
we can rearrange the formula to solve for h:
h = h0 + (V^2 * sin^2(θ)) / (2g)
Plugging in the known values, we get:
h = 0.9 + (41.7^2 * sin^2(44)) / (2 * 9.8)
Calculating this, we find:
h ≈ 7.9 m
Therefore, the ball would be approximately 7.9 m (or 26 ft) above the 3.0 m (10 ft) high fence if the fence was 116 m (380 ft) from home plate.
Part A:
To solve this problem, we can break it down into horizontal and vertical components.
Horizontal Component:
The horizontal component of the ball's initial velocity remains constant throughout its flight because there is no horizontal acceleration. We can calculate the horizontal distance traveled by the ball using the formula:
horizontal distance = horizontal component of velocity × time
Since the initial velocity is 44° above the horizontal, the horizontal component of velocity can be found using trigonometry:
horizontal component of velocity = initial velocity × cos(44°)
Vertical Component:
The vertical component of the ball's initial velocity changes due to the effect of gravity. The initial vertical velocity can be found using trigonometry:
initial vertical velocity = initial velocity × sin(44°)
Once we have the initial vertical velocity, we can use the kinematic equation to find the time of flight:
vertical distance = initial vertical velocity × time - 0.5 × acceleration due to gravity × time^2
where the vertical distance is 0.9 m (3.0 ft) and acceleration due to gravity is 9.8 m/s^2.
Solving for time, we get:
0.9 = (initial vertical velocity) × time - 0.5 × 9.8 × time^2
After finding the time, we can calculate the final horizontal distance by substituting it into the horizontal distance formula.
To produce a home run of 188 m (618 ft), the final horizontal distance should be 188 m. So, we can set up the equation:
final horizontal distance = (horizontal component of velocity) × time
Solving these equations simultaneously will give us the initial speed of the ball.
Part B:
To determine how high the ball would be above a fence 3.0 m (10 ft) high if the fence is 116 m (380 ft) from home plate, we can use similar calculations.
We can calculate the time it takes for the ball to reach the fence horizontally using the equation:
time = horizontal distance / (horizontal component of velocity)
Then, using the time calculated, we can find the height above the fence using the formula:
vertical distance above the fence = (initial vertical velocity) × time - 0.5 × acceleration due to gravity × time^2
Substituting the known values, we can find the height above the fence.