Finding the initial speed of the baseball?

[gap0063]

1. A home run is hit such a way that the baseballjust clears a wall 27 m high located 147 mfrom home plate. The ball is hit at an angleof 37◦ to the horizontal, and air resistance isnegligible. Assume the ball is hit at a heightof 1 m above the ground.The acceleration of gravity is 9.8 m/s2 .What is the initial speed of the ball?Answer in units of m/s.



2. tried: xf=xi+vxit+.5g^2
tan theta= vy/vx => vy= vx*tan theta
yf=vyt+.5gt^2



3. with the above equation I got:
147= 0+ vxi(4.30407) + 4.9(4.30407)^2=13.0638 wrong
I got t= 4.30407 by pluging in vy= vx*tan theta into yf=vyt+.5gt^2

 

[Delphi51]

I got t= 4.30407 by plugging in vy= vx*tan theta into yf=vyt+.5gt^2

I don't understand how you got the time. Surely you still have two unknowns, time and vx or vy? I got a slightly different answer for the time.

This problem yields to the old high school routine of making two headings for "horizontal" and "vertical", writing d = vt for the horizontal and d = vt + ½at² for the vertical. I would use v*cos(37) for the horizontal v, and v*sin(37) for the vertical initial v. This way you have two equations and two unknowns (v and t).