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February 28, 2015

February 28, 2015

Posted by **steve** on Friday, September 12, 2008 at 8:19pm.

- physics -
**bobpursley**, Friday, September 12, 2008 at 8:54pmYou have two dimensions to work with: horizontal and vertical

Vertical:

hfinal=hinitial+35.5sin49.9*t -1/2 g t^2

let hinitial be zero, so hfinal is .877.

You can solve for time here, using the quadratic equation.

Horizontal:

distance=35.5cos49.9 t solve the is for distance the ball went.

Finally, the distance the ball player ran is the above -116.

avgspeed=distaneballplayerran/time

- physics -
**GK**, Friday, September 12, 2008 at 9:04pmThe initial vertical component of the velocity is:

Vyo = (35.5m/s)(sin49.9) = _____ m/s

The vertical distance can be described by:

y = (Vyo)t + (1/2)(-9.8m/s^2)t^2

At the time the ball was caught:

0.877m = (Vyo)t + (1/2)(-9.8m/s^2)t^2

Substitute the value for Vyo and rearrange into a quadratic equation in standard quadratic trinomial form.

Solve for the time, t and use the larger of the two values.

The center fielder’s speed is (116m) / t = _____ m/s

- physics -
**bobpursley**, Friday, September 12, 2008 at 9:26pmThanks, George.

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