Stuck on Projectile Motion question

  • Thread starter coldjeanz
  • Start date
  • #1
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Homework Statement



There's a cannon sitting at the edge at the top of a 350 m cliff that fires a 15kg projectile at a muzzle velocity of 250 m/s. On the opposite side there's another cliff that is 470 m in height from the bottom of the canyon as well and has a target on it.

So the question is if you fire the projectile out at an angle of 30 degrees from the horizontal and it hits the target on the other side, what is the horizontal distance to that target?


Homework Equations



xf-xi (d) = vox (t) + 1/2a(t)^2

yf - yi = voy(t) + 1/2a(t)^2


The Attempt at a Solution




What I have tried to do:

For the first one I ended up with d = vox(t) because the other junk will go to 0 since there is no acceleration in the x direction.

For the second formula I attempted to solve for t so that I could plug that back in for t in the first formula to get the distance.

So I did

470m - 350m = 0 + 1/2(-9.8m/s^2)(t)^2

I solved for t here and plugged it back into the t in the first formula but it's not giving me the correct answer. The correct answer is 5300m but I'm getting something around 1000. Can someone guide me here and point out what I'm doing wrong?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
993
13
... 470m - 350m = 0 + 1/2(-9.8m/s^2)(t)^2 ...
The only mistake is that the INITIAL velocity in the vertical direction is not 0 but a component of 250m/s.
 

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