Master Projectile Motion: Find Maximum Height with Initial Speed and Angle

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To find the maximum height of a projectile launched at an angle theta with an initial speed v0, the time to reach maximum height (tmax) is calculated as tmax = v0sin(theta)/g. The total time of flight (tr) is twice this value, and the height at tmax can be expressed as H = y(tmax). The vertical velocity component is v0sin(theta), which allows for determining the distance traveled over time. To prove that one quarter of the horizontal distance (deltaX) times the tangent of theta equals the maximum height (deltaY), further clarification and mathematical derivation are needed.
Kalie
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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 I am not sure how to do that...I mean were am I able to plug those in so that I can get the answer
 
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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?
 
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