Posted by **Genevieve** on Friday, September 17, 2010 at 4:19pm.

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

xf=xi+vxit+.5g^2

tan theta= vy/vx => vy= vx*tan theta

yf=vyt+.5gt^2

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

- Physics- Mechaincs -
**bobpursley**, Friday, September 17, 2010 at 4:40pm
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.

- Physics- Mechaincs -
**Damon**, Friday, September 17, 2010 at 4:41pm
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

- Physics- Mechaincs -
**bobpursley**, Friday, September 17, 2010 at 4:41pm
and, on the yf, don't forget the initial height.

Yf=yi+Vy*t-1/2 g t^2

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