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raymondjg · Aug 27, 2010, 04:05 PM

Batted ball initial velocity of 105 mph. Ball hit waist high at 3 ft off ground. "Line drive" ball clears fence 350 ft away at 25 ft altitude. How far did it go? Disregard weather.

How far would it go at 110 mph and 115 mph initial velocity?

Thanks, Raymond

Unknown008 · Aug 28, 2010, 01:03 AM

You need to break the velocity of the ball into two components, one vertical and one horizontal.

On the vertical component, you have the acceleration due to gravity at 9.81 m/s^2.
One the horizontal component, you don't have any acceleration because air resistance is not considered.

Therefore, you get two equations:
(Let v be the initial velocity of the ball in metres)
$s_v = ut + \frac12 at^2 = v sin\theta + \frac12 (9.81)t^2 = v sin\theta + 4.905t^2$

$s_h = ut = v cos \theta t$

At some point in time, s_v = 25-3 = 22 ft and s_h = 350 ft.

Plug in these values to find the value of theta by eliminating t. Be careful about the units. Once you have the value of theta, find the time when the displacement = -3 ft using the first equation.

Using this time, find the horizontal distance covered by the ball using the second equation.

raymondjg · Aug 30, 2010, 11:21 AM

Thanks but I don't understand some of it. Can you please do the
Calculation and then I can see where the numbers are inserted.
Afterwards I can do it myself if I want to change the initial ball velocity. Raymond

Unknown008 · Aug 30, 2010, 11:40 AM

Okay.

I convert all to metres first.
22ft = 6.71 m
105 mph = 46.9 ms^-1
350 ft = 107 m

Using the first equation, I get:

$6.71 = 46.9t sin\theta - 4.905t^2$

Using the second equation, I get:

$107 = 46.9tcos\theta$

Using the second equation, I make t the subject of formula.

$t = \frac{107}{46.9cos\theta}$

$t = \frac{107}{46.9cos\theta}$

$t = \frac{2.28}{cos\theta}$

I now use that in the first equation:

$6.71 = 46.9 $\frac{2.28}{cos\theta}$ sin\theta - 4.905$\frac{2.28}{cos\theta}$^2$

This simplifies to:

$6.71 = 107 tan\theta-25.5 sec^2\theta$

$6.71 = 107 tan\theta-25.5 tan^2\theta - 25.5$

$25.5 tan^2\theta - 107 tan\theta + 32.2 = 0$

This gives

$tan\theta = 3.87$

So, theta = 75.5 degrees

This should be it if I didn't make any mistake in my calculations.

Now that you get this, use:

$-3 = 46.9tsin(75.5) - 4.905t^2$

To find t, that is the time that the ball takes to hit the ground.

Then, use that value of t in:

$s_h = 46.9tcos(75.5)$