- #1

Jaramia

- 15

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1. I am trying to derive the equation for maximum height of a trajectory. Must be in terms of v, theta, and g.

2.Vy=vsin(theta) dy=vsin(theta)t+0.5gt^2

3. I know that I must find the point when the vertical velocity is equal to zero and must find one equation for time. Then sub this equation into another.

So I found from the first equation that t=dysin(theta) and then substituted this equation to the second equation listed above. I then got this equation dy=vdsin(theta)+0.5g(dsin(theta))^2.

I don't know what to do from here because of the two displacements. And I don't even know what I am doing is correct as I can't solve.

Can anyone help lead me in the direct equation, it would also be helpful to acctually show me what the max height equation looks like.

2.Vy=vsin(theta) dy=vsin(theta)t+0.5gt^2

3. I know that I must find the point when the vertical velocity is equal to zero and must find one equation for time. Then sub this equation into another.

So I found from the first equation that t=dysin(theta) and then substituted this equation to the second equation listed above. I then got this equation dy=vdsin(theta)+0.5g(dsin(theta))^2.

I don't know what to do from here because of the two displacements. And I don't even know what I am doing is correct as I can't solve.

Can anyone help lead me in the direct equation, it would also be helpful to acctually show me what the max height equation looks like.

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