Posted by steve on Friday, September 12, 2008 at 8:19pm.
You 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
The 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
Thanks, George.
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