# Projectile motion & velocity on a certain height

1. Nov 16, 2005

### Lisa...

Hey!

I have the following problem to solve:

Consider a ball that is thrown with initial speed v0 at an angle a above the horizontal. If we consider it's speed at some height h above the ground show that v(h) is independent of a.

I started to solve the problem this way:

v= sqrt (vx^2 + vy^2)
vx= v0x= v0 cos a
vy= v0y- gt= v0 sin (a) -gt

Substitution of vx and vy in the first formula gives:

v= sqrt(v0^2 (cos^2)a + v0^2 ((sin^2)a) - 2 v0 gt sin a + g^2 t^2)
v^2= v0^2 (cos^2)a + v0^2 ((sin^2)a) - 2 v0 gt sin a + g^2 t^2
v^2= v0^2 ((cos^2)a + (sin^2)a) -2 v0 gt sin a + g^2 t^2
v^2= v0^2 -2 v0 gt sin a + g^2 t^2

Now I STILL have a term with a in it (-2 v0 gt sin a ) How in the world will I get rid of it?!

2. Nov 16, 2005

### Fermat

Use an eqn for Vy that is independent of t, but involves h

3. Nov 16, 2005

### Lisa...

But what would that equation be? I only know a formula for Vy with t... and what about Vx? That is always v0 cos a right? How will I get rid of that cosine?

4. Nov 16, 2005

### marlon

Well, you should use the equations for the position x and y. Can you write them down in general form ? Like you correctly did for the velocity in the x and y-direction.
Once you have these equations, you can use them to get rid of the angle-term (-2 v0 gt sin a ) in the velocity equation
marlon

Last edited: Nov 16, 2005
5. Nov 16, 2005

### Fermat

Hint: v² = u² + 2as

6. Nov 16, 2005

### Lisa...

You guys are really amazing! Thank you! I solved the problem the following way:
y(t)= h= y0 + v0y t + 1/2 gt^2 , with y0=0 (no initial height is given) and v0y= v0 sin a.

Therefore: h= v0 t sin a + 1/2 gt^2 and v0 t sin a= h- 1/2 gt^2.

Substitution in v^2= v0^2 -2 v0 gt sin a + g^2 t^2 gives:

v^2= v0^2 -2g(h-1/2gt^2) + g^2t^2 = v0^2 -2gh + g^2t^2 + g^2t^2= v0^2 -2gh + 2 g^2 t^2.

v= sqrt (v0^2 -2gh + 2 g^2 t^2.)

I can assume this is correct, right?

7. Nov 16, 2005

### Fermat

Just one sign error. The bolded bit above should be "- 1/2 gt²".

A simpler version could be,

V² = Vx² + Vy²

where Vx = Vo.cos@,
Voy = Vo.sin@,
Vy² = Voy² - 2gh (using the hint: v² = u² + 2as)
Vy² = Vo².sin²@ - 2gh

So,

V² = Vo².cos²@ + Vo².sin²@ - 2gh
V² = Vo²(cos²@ + sin²@) - 2gh
V² = Vo² - 2gh
===========

8. Nov 16, 2005

### Lisa...

Ah ok... how foolish of me! :surprised
Of course it needs to be -1/2gt^2