# Homework Help: Need Help Deriving Projectile Motion Equations

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1. Sep 16, 2016

### Rasiel

• Member advised to use the homework template and show their own efforts.
Could someone please assist me in deriving the following projectile motion equations from the following kinematics equations?

Kinematics Equations:
Δx = v0xt + ½at2 ; Δy = v0yt + ½at2
vx = v0x + at ; vy = v0y + at
Δx = ½(v0x + vx)t ; Δy = ½(v0y + vy)t
vx2 = v0x2 + 2aΔx ; vy2 = v0y2 + 2aΔy

Projectile Motion Equations:
Flight Time = [ 2v0sin(θ) ] / g
Max Height = [ v02sin2(θ) ] / 2g
Horizontal Range = [ v02sin(2θ) ] / g
Time to Reach Top = √((2 * MAX HEIGHT) / g)

2. Sep 16, 2016

### Ray Vickson

You need to show your own work, first. How much have you done so far? Where are you stuck?

3. Sep 16, 2016

### Rasiel

This isn't exactly a homework question. It's a question about general equations regarding projectile motion. I need to strengthen my understanding of them. Didn't know where to post non-homework related help.

4. Sep 16, 2016

### Staff: Mentor

Here at PF we treat "homework-like" questions the same way we treat assigned questions from a course of study. Usually such homework-like questions arise from a student wanting clarification on some matter associated with coursework and requires applying knowledge from such a course.

So this is in fact a good place for such questions, but the forum rules apply.

Why don't you pick one or more of the items form your list an make an attempt? I think that once you get through one or two with help you'll find that you can be successful with the rest.

5. Sep 17, 2016

### Rasiel

I have managed to derive the Max Height equation through the following process:

vy2 = v0y2 + 2ay
0 = v0y2 - 2gy : Since at the top of the trajectory vy must be 0.
v0y2 = 2gy
v02sin2θ = 2gy : v0sinθ is equivalent to the y-component of the velocity.
MAX HEIGHT = y = v0sin2θ / 2g

I think the next logical step would be to attempt to derive the Time to Top equation, I'll try my hand at it now, any advice is appreciated.

-EDIT-

Here is where I've arrived so far with respect to the Time to Top equation:

y = v0yt - ½gt2
v0sin2θ / 2g = v0sinθ - ½gt2
v0sin2θ / g = 2v0sinθ - gt2
(v02sin2θ / g2) - (v0sinθ / g) = t2
½gt2 = MAX HEIGHT - v0y

I can't seem to see where to go from here though...

-EDIT- Realized a mistake was made.

-EDIT-

Here is my new attempt:

y = v0yt - ½gt2
v02sin2θ / 2g = v0sinθt - ½gt2
gt2 - 2v0sinθt + 2H = 0

After applying the quadratic formula: t = 2v0sinθ / 2g = v0sinθ / g

Now I'm stuck :(.

-EDIT-

By way of simple algebraic manipulation:

t = v0sinθ / g
gt = v0sinθ
g2t2 = v02sin2θ
½gt2 = H
t2 = 2H / g
t = √(2H / g)

SUCCESS!

Last edited: Sep 17, 2016
6. Sep 17, 2016

### Rasiel

Can't seem to make any headway into the Horizontal Range and Flight Time equations, any guidance is appreciated. @gneill @Ray Vickson

7. Sep 17, 2016

### olivermsun

You have Time to Reach Top expressed as a function of MAX HEIGHT.

See if you can express Time to Reach Top directly as a function of $v_{0y}$ and $g$? If so that will get you on the right track.

8. Sep 17, 2016

### Rasiel

Time to Top = v0sinθ / g

So I guess by the parabolic nature of projectile motion we can just double this and get: Time of Flight = 2v0sinθ / g.

Then we can get Horizontal Range as follows:

Δx = v0cosθt - ½gt2
Δx = v0cosθ(2v0sinθ / g) - 0
Δx = 2v02sinθcosθ / g
Δx = 2v02sin(2θ) / g

Thanks for the assistant guys! It makes sense now.

The only question left in my mind regarding the topic is: How can these equations be extended to the scenario where a cannon is fired off a cliff and hits an area that is below parallel to the cannon. Or an instance where a cannonball is shot onto a cliff, above parallel to the cannon?

9. Sep 17, 2016

### Staff: Mentor

Some of the results won't be so nice, involving the solutions to quadratic equations for example. For the first case you have to add the initial height to your vertical equation of motion. Something like:

Δy = yo + v0yt + ½at2

while for the second case you assign a value to the final vertical displacement. Usually this leads to solving a quadratic equation at some point and you need to be careful to choose the solution that fits the scenario (imaginary roots are not physical, but there may be two real roots that are candidates --- consider that for a parabolic trajectory the projectile can go through every y value twice, once on the way up and once on the way down).