Projectile Help

1. May 30, 2010

Motorboar

1. The problem statement, all variables and given/known data
According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor-league game. The ball traveled 188m (618 ft.) before landing on the ground outside the ballpark.

Assuming that the ball's initial velocity was 40.0 degrees above the horizontal, and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.900 m (3.00ft ) above ground level? Assume that the ground was perfectly flat.

How far would the ball be above a fence 3.00 m (10.0 ft) in height if the fence were 116 m (380ft ) from home plate?

2. Relevant equations
v0x=v0cos(40)
v0y=v0sin(40)
x-x0=v0xt
y-y0=v0yt+1/2at^2
ay=-o.81
y-y0=-0.9
ax=0
x-x0=188

3. The attempt at a solution
-0.9 =v0sin40t + 1/2(-9.81)t^2
t= 188/v0cos(40)
0 = (v0sin40)(188/v0cos(40))-(4.9)(188/v0cos(40))^2+0.9
graphed it on a calculator to avoid messy algebra and got the wrong answer of 32.97.
No Idea why this is wrong. Any ideas?

2. May 30, 2010

pgardn

I got 43.1 m/s... I just went with your numbers. I would try it again.

3. May 30, 2010

Motorboar

43.1 is the right answer. Did you graph it? I been working with the input for like 30 minutes, do you mind telling me how you plugged it in?

4. May 30, 2010

pgardn

I just used your numbers. Maybe you squared the term for acceleration incorrectly?

-0.9 = sin40*(188/cos40) - 4.9*(188/Vo*cos40)^2

The Vo cancels out of the first term and the Vo in the second term gets squared so you eventually have to take the square root. It a bit of a pain as you said.

5. May 30, 2010

Motorboar

If im not lazy and work it out by hand, I get it. Thanks for your help.