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?

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

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

Thank you! It made sense! :D

To find the angle of elevation at which the ball was thrown, we can rearrange the equation:

d = v^2/(32) * sin(2theta)

In this case, we have:

d = 200 feet
v = 75 miles per hour

First, let's convert the velocity from miles per hour to feet per second. There are 5280 feet in a mile and 3600 seconds in an hour, so we have:

v = 75 * 5280 / 3600 = 110 feet per second

Now, we substitute the given values into the equation and solve for theta:

200 = (110^2)/(32) * sin(2theta)

Simplifying further:

200 = 12100/32 * sin(2theta)

200 = 378.125 * sin (2theta)

Dividing both sides by 378.125:

0.529 = sin(2theta)

To find the angle, we need to take the inverse sinus of both sides:

sin^(-1)(0.529) = 2theta

Using a calculator, we find:

2theta ≈ 31.7 degrees

Now divide by 2 to get the angle of elevation:

theta ≈ 15.85 degrees

So, the angle of elevation at which the ball was thrown is approximately 15.85 degrees. The book may have rounded to 16.0 degrees or 74.0 degrees, which are slightly different but still within an acceptable range of approximation.