# Precalculus

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When a projectile leaves a starting point at an angle of elevation of theta with a velocity v, the horizontal distance it travels is determined by:

d=v^2/(32) sin2theta

Where d is measured in feet and v in feet per second.

An outfielder throws the ball at a speed of 75 miles per hour to the catcher who is 200 feet away. At what angle of elevation was the ball thrown?

• Precalculus -

sub in the values
200 = 75^2/2 sin^2 Ø
400/5625 = sin^2 Ø
sin Ø = √(400/5625) = 20/75 = 1/3
Ø = 19.5°

• Precalculus -

The answer says:16.0 or 74.0 degrees :|

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