Need Help Deriving Projectile Motion Equations

[Rasiel]

Could someone please assist me in deriving the following projectile motion equations from the following kinematics equations?

Kinematics Equations:
Δx = v_0xt + ½at² ; Δy = v_0yt + ½at²
v_x = v_0x + at ; v_y = v_0y + at
Δx = ½(v_0x + v_x)t ; Δy = ½(v_0y + v_y)t
v_x² = v_0x² + 2aΔx ; v_y² = v_0y² + 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)

 

[Ray Vickson]

Could someone please assist me in deriving the following projectile motion equations from the following kinematics equations?

Kinematics Equations:
Δx = v_0xt + ½at² ; Δy = v_0yt + ½at²
v_x = v_0x + at ; v_y = v_0y + at
Δx = ½(v_0x + v_x)t ; Δy = ½(v_0y + v_y)t
v_x² = v_0x² + 2aΔx ; v_y² = v_0y² + 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)

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

 

[Rasiel]

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

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.

 

[gneill]

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.

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.

 

[Rasiel]

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

v_y² = v_0y² + 2ay
0 = v_0y² - 2gy : Since at the top of the trajectory v_y must be 0.
v_0y² = 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 = v_0yt - ½gt²
v₀sin²θ / 2g = v₀sinθ - ½gt²
v₀sin²θ / g = 2v₀sinθ - gt²
(v₀²sin²θ / g²) - (v₀sinθ / g) = t²
½gt² = MAX HEIGHT - v_0y

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 = v_0yt - 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!

 

[Rasiel]

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

 

[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.

 

[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?

 

[gneill]

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?

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 = y_o + v_0yt + 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).