How Far Does a Projectile Travel When Fired from a Cliff?

[roe]

<b>Q:</b>A projectile is shot from the edge of a vertical cliff 60.0 m above the ocean. It has a speed of 100 m/s and is fired at an angle of 35.0 degrees below the horizon. How far from the foot of the vertical cliff does the projectile hit the water?

My take on the problem:

          __________
          |\ ) 35 degree angle
          | \      
          |  \
          |   \
          |    \ 
          |     \
          |______\
  ^          ^ what I'm trying to find (xf)
60m

So what I'm trying to do is figure out how far this projectile will go in the +x direction before it reaches the bottom of the cliff (60 m).

My problem is that I'm getting quite confused as to how to start solving this problem. Please help me get started!

 

[whozum]

The first thing you want to find is how long it takes to hit the ground. From there you'll be able to find everything else. Don't forget that it has an initial vertical velocity..

 

[twofish]

Hi Roe,
I would start with getting the inside angle, so 90-35. That gives you two angles and one side (60M).

 

[roe]

OK, I'm not seeing how to determine when it hits the ground. I've been doing 1-dimensional problems for the last couple of hours but I'm really not understanding how to set this problem up. What I have so far is:

Vxi = (100 m/s) cos 55 degrees = 57.36 m/s
Vyi = (100 m/s) sin 55 degrees = 81.92 m/s

yf = yi +vyi(t) + .5(ay)(t^2)

with yi = 0, yf = -60m, ay = -g, Vyi = 81.92 m/s

so...

-60m = (81.92m/s)(t) - .5(-9.8m/s/s)(t^2)

And then I'd solve the quadratic equation for t.

Am I on the right track with this?

 

[twofish]

Hi Roe,
I don't see the time making a diff at all.
You have 3 angles (all) and one side, the rest should just be a matter of trig.
Check out http://www.math2.org/math/trig/identities.htm