Abatter hits a pitched ball when the center of the ball is 1.21 m above the ground. the ball leaves the bat at an angle of 45° with the ground. with that launch, the ball should have a horizontal range (returning to the launch level) of 116 m. (a) does the ball clear a 8.79-m-high fence that is 106 m horizontally from the launch point? (b) at the fence, what is the distance between the fence top and the ball center?

a) The ball will clear the fence.

b) The distance between the ball and the top of the fence is 1.41 m.

Explanation:

The equation for the position of the ball is given by the vector "r":

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector of the ball at time t

v0 = initial velocity

x0 = initial horizontal position

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point.

When the ball reaches a horizontal distance of 116 m, the y-component of its vector position (r final in the figure, in red) is 0. Then:

r final = (116 m , 0)

Using the equations for the x and y-component of "r" we can calculate the total time of flight and the initial velocity:

x = x0 + v0 · t · cos α

116 m = 0 m + v0 · t · cos 45°

116 m / cos 45° · t = v0

Replacing v0 in the equation of the y-component, we can obtain the final time of flight:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 m = 0 m + 116 m / (cos 45° · t) · t · sin 45° - 1/2 · 9.8 m/s² · t²

0 m = 116 m - 4.9 m/s² · t²

-116 m /  - 4.9 m/s² = t²

t = 4.87 s

Then, the initial velocity will be:

v0 =  116 m / cos 45° · t

v0 = 116 / cos 45° · 4.87 s

v0 = 33.7 m/s

a) Now, let´s find the y-component of the vector position when the x-component is 106 m (vector r in the figure). If the y-component plus 1.21 m is greater than 8.79, then, the ball will clear the fence.

Let´s find the time at which the position vector of the ball has an x-component of 106 m.

x = x0 + v0 · t · cos α

106 m = 0 m + 33.7 m/s · t · cos 45°

106 m / 33.7 m/s · cos 45° = t

t = 4.45 s

The y-component at that time will be:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 0 m + 33.7 m/s · 4.45 s · sin 45° - 1/2 · 9.8 m/s² · (4.45 s)²

y = 9.01 m

Then, the height of the ball relative to the ground when the horizontal distance is 106 m (where the fence is) is 9.01 m + 1.21 m = 10.2 m. The ball will clear the fence.

b) The distance between the ball and the fence will be the height of the ball relative to the ground minus the height of the fence:

distance = 10.2 m - 8.79 m = 1.41 m

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