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:

Question

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?

-your answer was 19.5 degree, however, the book says 16.0 or 74.0 degrees

Answer 1

I totally messed up that question
3 errors:
1. did not notice that v was mph and distance was feet
2. wrote down division by 2 instead of 32
3. read sin 2Ø as sin^2 Ø

so let's try that again .....
75 miles/hr = 75(5280)/3600 ft/sec
= 110 ft/sec

then 200 = 110^2 / 32 (sin 2Ø)
sin 2Ø = .5289256..
2Ø = 31.93 or 2Ø = 148.067
Ø = 15.966 or 16°
or
Ø = 74.03 or 74°

sorry about that

Answer 2

Thank you! It made sense! :D