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## Homework Statement

The center of a baseball of diameter 8.20 cm is 1.320 m vertically above the plate when it is hit. The blast sends it off at an angle of 30.5° above the horizontal with an unknown initial speed. The outfield fence is 94.0 m away and 11.20 m tall: the ball just clears it. Ignoring aerodynamic effects, what was the initial speed of the baseball?

## Homework Equations

-d=vt for the x direction

-any free fall equation for the y direction

## The Attempt at a Solution

dx= 94.0m dy= 9.88m (subtracted the height in which the ball was hit at)

Vx= vcos(30.5 Viy=?

t= 94.0/vcos(30.5 a= -9.8m.s^2

d=vit+1/2at^2

9.88=vi(94.0/vcos(30.5)-4.9(94.0/vcos(30.5)^2

v^2(-99.22)= -58323.769

v^2= 5884.13

v=76.7m/s

where am i going wrong in this problem? help will be greatly appreciated, thanks!