How Do You Calculate the Maximum Height of a Projectile Using v, Theta, and g?

[Jaramia]

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.

 

[hotcommodity]

You're going to need more than one equation to figure this out, but I'll narrow it down for you. We're given that the final velocity v_y is equal to zero, and we're given g, v, and theta. We want to begin by placing the initial velocity in the x-direction and the initial velocity in the y-direction in terms of v and theta. To do so we use the equations:

$$v_{0x} = v cos \theta$$

$$v_{0y} = v sin \theta$$

From the kinematic equations, we'll want to use:

$$y = v_{0y} - 0.5(g)t^2$$

You know how to get v _0y in terms of v and theta, so now we focus on how to get the time t into the given terms. Note again that the final velocity in the y-direction is zero, thus:

$$0 = v_{0y} -gt$$

Solve this for t and you're in business.

 

[Jaramia]

In the kinematics equation, what happened to the t?
and can you explain what Voy is opposed to Vy?

 

[hotcommodity]

Oh sorry you're right, that should be:

$$y = v_{0y}t - 0.5(g)t^2$$

Also, v_0y is the initial velocity in the y-direction, and v_y is the final velocity in the y-direction, which should be zero, right?

I forgot to put the "v" in the initial velocity equations the first time around. Just to clarify, they should be:

$$v_{0x} = v sin \theta$$

$$v_{0y} = v cos \theta$$

Does this make sense?

 

[Jaramia]

can you possibly explain how Voy-gt equals zero?

 

[hotcommodity]

One of the kinematic equations for the y-direction is $$v_y = v_{0y} -gt$$. If we know the final velocity in the y-direction to be zero, then we have:

$$0 = v_{0y} -gt$$

Does that make sense?

 

[blochwave]

That would be 0=Voy-g*tfinal then

In general 0=Voy-gt is not true, except at tfinal, which is the time when it reaches the peak

 

[Jaramia]

ok everything makes sense now.
i substituted the equation and came up to
dy=(v^2cos(theta)^2)/g) - (vcos(theta)/2)
Have i done this right so far? if so, how can i continue?

 

[hotcommodity]

That would have been right if I had got more than three hours of sleep, I do apologize. It should be sine of theta, not cosine in your equation. You'd use:

v0x = vcos \theta

v0y = v sin \theta

But substitute the sin in for the cos, and you're good to go.

 

[hotcommodity]

Oops, one last edit. You're missing one term, and a few squares. Your answer should be:

dy=(v^2sin(theta)^2)/g) - (v^2sin(theta)^2/g2)

 

[Jaramia]

ok thank!

 

[hotcommodity]

You're welcome, but also note that the final answer can be simplified :)

 

[Jaramia]

ya i got it, thanks again