Answer:
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
