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

In summary: No problem!In summary, to find the equation for maximum height of a trajectory in terms of initial velocity (v), launch angle (theta), and acceleration due to gravity (g), one must use the equations v_{0x} = v cos \theta and v_{0y} = v sin \theta to find the initial velocity components, and then use the kinematic equation y = v_{0y}t - 0.5(g)t^2 with the given information of final velocity in the y-direction being zero and the initial velocity components in terms of v and theta. This will result in the equation dy=(v^2sin(theta)^2)/g) - (v^2sin(theta)^2/g2), which
  • #1
Jaramia
15
0
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.
 
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  • #2
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 vy 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:

[tex] v_{0x} = v cos \theta [/tex]

[tex] v_{0y} = v sin \theta [/tex]

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

[tex] y = v_{0y} - 0.5(g)t^2 [/tex]

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:

[tex] 0 = v_{0y} -gt [/tex]

Solve this for t and you're in business.
 
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  • #3
In the kinematics equation, what happened to the t?
and can you explain what Voy is opposed to Vy?
 
  • #4
Oh sorry you're right, that should be:

[tex] y = v_{0y}t - 0.5(g)t^2 [/tex]

Also, v0y is the initial velocity in the y-direction, and vy 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:

[tex] v_{0x} = v sin \theta [/tex]

[tex] v_{0y} = v cos \theta [/tex]

Does this make sense?
 
  • #5
can you possibly explain how Voy-gt equals zero?
 
  • #6
One of the kinematic equations for the y-direction is [tex] v_y = v_{0y} -gt [/tex]. If we know the final velocity in the y-direction to be zero, then we have:

[tex] 0 = v_{0y} -gt [/tex]

Does that make sense?
 
  • #7
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
 
  • #8
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?
 
  • #9
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.
 
  • #10
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)
 
  • #11
ok thank!
 
  • #12
You're welcome, but also note that the final answer can be simplified :)
 
  • #13
ya i got it, thanks again
 

1. What is the maximum height of trajectory?

The maximum height of trajectory, also known as the peak height or apogee, is the highest point reached by an object that is thrown, launched, or projected into the air. It is the point where the object stops moving upwards and begins to fall back towards the ground.

2. How is the maximum height of trajectory calculated?

The maximum height of trajectory can be calculated using the equation: h = (v02sin2θ) / (2g), where h is the maximum height, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (9.8 m/s2). This equation assumes no air resistance and a flat, horizontal surface.

3. What factors affect the maximum height of trajectory?

The maximum height of trajectory is affected by the initial velocity, launch angle, and acceleration due to gravity. Other factors such as air resistance, wind, and the shape of the object can also have an impact on the maximum height.

4. Does the maximum height of trajectory change if the object is launched from a higher or lower elevation?

Yes, the maximum height of trajectory will change if the object is launched from a higher or lower elevation. This is because the initial height affects the potential energy of the object, which then affects its maximum height. For example, launching an object from the top of a hill will result in a higher maximum height compared to launching the same object from ground level.

5. What is the relationship between the maximum height of trajectory and the range of the object?

The maximum height of trajectory and the range of the object are related in that they both depend on the initial velocity and launch angle. Generally, a higher maximum height will result in a longer range, as the object has more time to travel horizontally before reaching the ground. However, other factors such as air resistance and wind can also affect the range of the object.

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