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Kalie
Sep16-06, 01:46 PM
A projectile is fired from ground level at time 0, at an angle theta with respect to the horizontal. It has an initial speed v0. In this problem we are assuming that the ground is level.
Find , the maximum height attained by the projectile, okay it is just asking for the equation
I know I need the tmax
which is = v0sin(theta)/g
tr=2tmax
H=y(tmax)
so Height=y(tmax)
but Im not sure how to do that.......I mean were am I able to plug those in so that I can get the answer

Astronuc
Sep16-06, 02:35 PM
The vertical velocity is vosin\theta. If one knows the initial vertical velocity and rate of acceleration (or deceleration), then one can determine the distance traveled at any time.

It would be useful to show that tmax = v0sin(theta)/g.

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

Duncman
Sep17-06, 08:02 PM
I've been working many problems similar to this one and I think I'm getting good answers. The teacher through out a rough one that I could use some help with though.

I'm supposed to prove that one quarter the distance the projectile travels times the tangent of theta is how high the projectile goes. I can see that this is true, but I'm not sure how to prove it.

Its expressed as deltaY = .25 deltaXtantheta

Any ideas?