Question

On a baseball field, the pitchers mound is 60.5 feet from home plate. During practice, a batter hits a ball 186 feet at an angle of 40 degrees to the right of the pitchers mound. An outfielder catches the ball and throws it to the pitcher. Approximately how far does the outfielder throw the ball?!

Answer 1

The outfielder threw the ball 145 feet away.

Explanation:

We are given that,

Distance of mound from home plate = 60.5 feet

Distance of outfielder from home plate = 186 feet

Angle made by the batter = 40°

So, using the cosine law, we have,

`c^(2)=a^(2)+b^(2)-2ab\cos \theta`

i.e.
`c^(2)=(60.5)^(2)+(186)^(2)-2* 60.5* 186* \cos 40`

i.e.
`c^(2)=38256.25 -22506* 0.766`

i.e.
`c^(2)=38256.25 -17239.596`

i.e.
`c^(2)=21016.654`

i.e. c = ±144.97 ≈ ±145

Since, the distance cannot be negative.

Thus, the outfielder threw the ball 145 feet away.

Answer 2

Use cosine law to solve for x

`x^2=60.5^2+186^2-2*60.5*186* cos(40^o) \\x^2=38,256.25-17,240.6 \\x^2=21,015.65 \\x= √(21,015.65) \\x=145`