posted by Genevieve .
A home run is hit such a way that the baseball
just clears a wall 27 m high located 147 m
from home plate. The ball is hit at an angle
of 37◦ to the horizontal, and air resistance is
negligible. Assume the ball is hit at a height
of 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.
I tried using these equations
tan theta= vy/vx => vy= vx*tan theta
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
1. The problem statement, all variables and given/known data
147= 0+ vxi(4.30407) + 4.9(4.30407)^2
well, I have issues with this. Why is gravity in the horizontal equation? (Besides,if it were, it would be negative).
In the yf=vyt+.5gt^2 , the last term should be negative (-.5gt^2). Upwards is postive, downward negative.
let V = initial velocity
xf =xi + vxi t
There is no acceleration in the horizontal direction, horizontal speed is constant!
vxi = V cos 37
so xf = 0 + .8 V t = 147
solve for V t
Now do y direction (up)
yf = yi + V t sin 37 - .5 (9.8) t^2
Note g is - if up is +
27 = 1 + V t sin 37 -4.9 t^2
we know V t
solve for t
and, on the yf, don't forget the initial height.
Yf=yi+Vy*t-1/2 g t^2