projectile motion

[Monsu]

hey, pls smne take a look at this question, i need help.
" A man stands on a cliff 15m above the ground and throws a little rock off the cliff with a velocity of 30m/s at an angle of 33.0degrees. Air resistance can b neglected. what is the speed of the rock just before it hits the ground?"
i guessed the equation i'd use to solve this would be " v= u^2 + 2gh " but then, what would b the intial velocity? the height would b the height of the cliff + the height above the cliff, right? then since at that point , the rock is at maximum height, would the velocity at that point be 0 ?
thanks!!! :smile:

[Pyrrhus]

At Max Height Vy = 0, Vy = VoSin(Angle), you can find the speed by knowing Y=0 when the rock hits the ground

[Pyrrhus]

i guessed the equation i'd use to solve this would be " v= u^2 + 2gh " but then, what would b the intial velocity? the height would b the height of the cliff + the height above the cliff, right? then since at that point , the rock is at maximum height, would the velocity at that point be 0 ?

"v= u^2 + 2gh", i'm not familiar with that equation but it looks like V^2 = Vo^2 +2a(Y-Yo), if so u will be initial speed, and V must be squared, also pick a sign convention, i mean if up and right is positive then left and down is negative, so g will be negative.

[Monsu]

yeah, i meant v^2 = u^2 + 2gh, so i guess what u are saying is that in this case u will b 0, making u^2 = 0. therefore i am left with v^2 = 2gh , i'll take g as negative, since the ball is falling. right?

[Doc Al]

If you are starting your motion at the top of the cliff, then u is not zero. If you take the top of the cliff as your starting point, and choose down to be negative, then both the acceleration and h are negative.

On the other hand, if you take the top of the trajectory as your zero point, then u is zero. But then you have to find how high it goes. Why waste the time?

[Monsu]

the first part of the question was to find how high above the cliff the rock went. and then to find the speed just before it hit the ground. also, how would i find the horizontal distance? isn't that the range? R = [U^2 sin2(theta)] / g ?

[Doc Al]

also, how would i find the horizontal distance?
Find the time it takes for the rock to hit the ground. You know the horizontal speed of the rock, so find the horizontal distance.

The range formula only works for level ground, not off a cliff.

[Pyrrhus]

Monsu a tip about all the formulas in Projectile Motion... The only formulas you'll ever need are the kinematic equations