# Need help with trajectory question please

## Homework Statement

You have been employed by the local circus to plan their human cannonball performace. For this act, a spring-loaded cannon will shoot a human projectile, the Great Flyinski, across the big top to a ew below. The net is located 5.0 m lower than the muzzle of the cannon from which the Great Flyinski is launched. The cannon wil shoot the Great Flyinski at an angle of 35.0° above the horizontal and at a speed of 18.0 m/s. The ringmaster has asked that you decide how far from the cannon to place the net so that hte Great Flyinski will land in the net and not be splattered on the floor, which would greatly distrub the audience. What do you tell the ringmaster? So basically the question is asking to find the final distance in the x direction.

## Homework Equations

vf = vi + at

xf-xi = vit + 1/2at^2

vf^2 = vi^2 + 2a(xf - xi)

xf - xi = 1/2(vf + vi)t

## The Attempt at a Solution

We tried finding the final velocity on the y, then use that to find the time, then use the time to find the x (the final distance). But the problem is when using:

vf^2 = vi^2 + 2a(xf - xi)

I got...

vf^2 = (10.3 m/s)^2 + 2(-9.8 m/s^2)(0 m - 5 m)

10.3 is found by doing 18.0cos35° and 0 m and -5 m comes from the locations of the cannon and the net (net is 5 m below cannon).

so I got vf (for y) = 14.3 m/s

then I did time

yf - yi = 1/2(vf + vi)t

0 - 5 m = 1/2(14.3 m/s + 10.3 m/s)(t) = .41 s

tried pluggin time and final v for y in...

xf - xi = 1/2(vf + vi)t

xf - 0 = 1/2(14.7 m/s + 14.7 m/s)(t)

vf and vi for x are the same because acceleration along the x axis doesn't change.

... = 6.0 m (which I got for how far away the net is from where the cannon shoots)

The correct answer is actually 37.1 m, can someone explain why what we did would not work?

gneill
Mentor
Be careful that you don't lose directional information (the sign) when you use formulae with squares. There may be one or more velocities that you've determined as going in the positive direction that should have been negative....

verty
Homework Helper
One thing I noticed, look again at 18 cos 35°. The adjacent side is horizontal, not vertical.

hey guys, thanks. the cos thing was a misprint (I did it correctly on paper), although good catch!

Vertigo, you are correct! I went over it again and noticed that final velocity on 'y' should be negative, but I came up with a positive number when I unsquared 204.09. So that is pretty tricky, but I figure as long as I remember that final y velocity is always negative (if projection is going DOWN), then that shouldn't be too bad.

Thanks alot you guys were a big help, this %*&@ was bugging me for a while, lol.