How far would the ball be above a fence 3.0 m (10 ft) high if the fence was 116 m (380 ft) from home plate?

Let V be the initial speed of the ball. Because the ball was hit angle an angle, there are two speed components: one is in x-direction, and the other one is in the y-inderect. The ball travels at a constant speed horizonally but accelerates vertically.

Break down the intial speed

Vx = Vcos45

Vy = Vsin45

find the time it takes the ball to hit the ground 116m away horizonally

x = vt

116 = Vcos45 * t

t = 116 / Vcos45

Xf = .5at^2 + Vt + Xi

Xf = final position (0m)

Xi = initial position (.9m)

a = acceleration

t = time

V = speed in y-derection

we just found the time, which is 188 / Vcos45, plug it in

0 = .5(-9.8)(188 / Vcos45)^2 + Vsin45 * (188 / Vcos45) + .9

-.9 = -4.9(188 / Vcos45)^2 + 188

-188.9 = -4.9(188 / Vcos45)^2

38.551 = (188 / Vcos45)^2

6.208 = 188 / Vcos45

6.208 * Vcos45 = 188

V = 188 / (6.208cos45)

V = 42.82 m/s

The ball was hit with the speed of 42.82m/s

Find the it takes the ball to travel 116m

x = vt

116 = 42.82cos45 * t

t = 3.831s

Xf = .5at^2 + Vt + Xi

Xf = .5(-9.8)(3.831)^2 + 42.82sin45 (3.831) + .9

Xf = 45m

the ball was 45 - 3 = 42m above the fence

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