SOLUTION: Hello! I'm so so confused on this problem for my Trig class. My dad attempted to help me solve it (I'll include the work below, though it is probably wrong) but his method does n

Algebra ->  Trigonometry-basics -> SOLUTION: Hello! I'm so so confused on this problem for my Trig class. My dad attempted to help me solve it (I'll include the work below, though it is probably wrong) but his method does n      Log On


   

Question 1179259: Hello! I'm so so confused on this problem for my Trig class. My dad
attempted to help me solve it (I'll include the work below, though it is
probably wrong) but his method does not seem right to me. Thanks!
"Determine the angle required for a baseball to leave the bat with an
initial velocity of 90 miles per hour and first hit the ground 300 feet
horizontally from home plate if contact is made 3 feet above the ground."
300=%28132+cos+theta%29t, 0=-16t%5E2%2B%28132+sin+theta%29t%2B3
300%28132+cos+theta%29=t →
-3=-16%28300132+cos+theta%292%2B%28300+tan+theta%29
Answer: 37 degrees
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The vertical component of feet above the ground

h%5BF%5D=h%5B0%5D%2Bv%5B0%5Dt-16t%5E2

h0 = 3     <--initial height

hF = 0     <--final height

90 mi/h = 90*(88/60) ft/s = 132 ft/s

v0 = 132sin(θ)
 
0=3%2B132sin%28theta%29t-16t%5E2

---------------------------------

The horizontal component of distance along the ground.
(Think of this as the shadow of the ball moving across the ground at
noontime with the sun directly overhead. The shadow moves along the
ground at a constant velocity.)

x%5BF%5D=x%5B0%5D%2Bvt

x0 = 0     <--initial distance along ground

xF = 300     <--final distance along ground

90 mi/h = 90*(88/60) ft/s = 132 ft/s   <-- same as above

v = 132cos(θ)
 
300=0%2B132cos%28theta%29t

300=132cos%28theta%29t

-----------------------------

So we have this system of equations to solve:

system%280=3%2B132sin%28theta%29t-16t%5E2%2C300=132cos%28theta%29t%29

Solve the second equation for t and substitute in the first
and get θ ≈ 16.08725911o   <--answer

Substitute that back and get

t ≈ 2.36535260 seconds  <-- you weren't asked for the time

I used a TI-84 calculator to get the answer

Round the answers as you were instructed.

I think your dad was on the right track, but the work is
so complicated, it's easy to make a mistake.  Maybe I made
a mistake. 

Edwin