What is the min angle from the horizontal (as the ball leaves the bat) required for a hit to be a home run?

Question

What is known:

The outfield fence is 10 feet high from foul line to foul line.

The outfield fence is 400 feet from home plate.

A ball is hit 3 feet above the plate.

The ball leaves the bat at a rate of speed 100mph.

I forgot.
Neglect effects of wind.

Answer 1

Been awhile since physics but...

First convert 100mph to feet/s so you're working with the same units..146.666667 ft/s. And using kinematics we obtain the following parametric equations (where S is initial position)

x(t) = Vi cos(Ө)*t + S
y(t) = -1/2gt^2 + sin(Ө)*t + S
So,
x(t) = 146.666667 cos(Ө)*t
y(t) = -16t^2 + 146.666667 sin(Ө)*t+3
initial height is 3 and a=-32ft/s^2

At the the minimum angle.. y will be exactly 10 feet (just passing over the fence) when x is 400

[1] 400 = 146.666667 cos(Ө)*t
[2] 10 = -16t^2 + 146.666667 sin(Ө)*t+3

solving [1] for t and subbing into [2]
t = 400/[146.666667 cos(Ө)]

10 = -16*(400/[146.666667 cos(Ө)])^2 + 400 tan(Ө) + 3
-16*(400/[146.666667 cos(Ө)]^)2 + 400 tan(Ө) - 7 = 0

Pretty messy, you need to solve for Ө somehow.. Probably best to graph this on a calculator and pick out the smallest zero

I ended up with a Ө of 19.3846308 degrees

Answer 2

this is physics stuff, just so you know. also it depends on air resistance also.

Answer 3

it is around 46degrees