A batted baseball leaves the bat at an angle of 25.0 above the horizontal and is

Question

A batted baseball leaves the bat at an angle of 25.0 above the horizontal and is caught by an outfielder 350 ft from home plate at the same height from which it left the bat.

Part A

What was the initial speed of the ball?

v = ____ m/s

Part B

How high does the ball rise above the point where it struck the bat?

H=_____m

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Explanation / Answer

the equation for the range is
R = v0^2/g*sin(2angle) --> plug in what you have given:
350 = v0^2/32*sin(2*25)
and solve for v0:
v0^2 = 350*32/sin(50)
v0 = 120.91 ft/s or 36.85 m/s

the maximum height is given by
height = v0^2*sin^2(angle)/(2g)
height = 120.91^2*sin^2(25)/(2*32)
max.height = 40.79 ft or 12.43 m

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