Asked by jake

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?

Answers

Answered by Reiny
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°
Answered by jake
The answer says:16.0 or 74.0 degrees :|
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