Need help with trajectory question please

[boringbum]

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]

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]

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

 

[boringbum]

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

 

[rwisz]

First of all, it is important to note that the equations you have listed are for motion in one dimension (either x or y), but the problem involves motion in both the x and y directions. Therefore, we need to use equations that take into account both the x and y components of the motion.

To solve this problem, we can use the following equations:

xf = xi + vixt + 1/2axt^2

yf = yi + viyt + 1/2ayt^2

where xf and yf are the final positions in the x and y directions, xi and yi are the initial positions, vix and viy are the initial velocities in the x and y directions, ax and ay are the accelerations in the x and y directions, and t is the time.

In this problem, we know the initial position (xi = 0 m) and the initial velocity in the x direction (vix = 18.0 m/s * cos(35°) = 14.7 m/s). We also know the initial position in the y direction (yi = 0 m) and the initial velocity in the y direction (viy = 18.0 m/s * sin(35°) = 10.3 m/s). The acceleration in the x direction is 0 m/s^2 because there is no force acting on the object in the x direction (neglecting air resistance). The acceleration in the y direction is -9.8 m/s^2 because the object is being pulled downward by gravity.

We also know that the final position in the y direction is -5.0 m (since the net is 5.0 m below the cannon) and we want to find the final position in the x direction (xf).

Plugging all of these values into the equations, we get:

-5.0 m = 0 m + (14.7 m/s)(t) + 1/2(0 m/s^2)(t^2)

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

Solving for t in the first equation, we get t = 0.34 s. Plugging this value into the second equation, we get xf = 5.0 m + (14.7 m/s)(0.34 s) +