# Projectile launch point problem

1. Oct 29, 2006

### SA32

I'm having some difficulty with a projectiles question.

"A projectile is fired with an initial speed of 26.0 m/s at an angle of 40.0 degrees above the horizontal. The object hits the ground 8.00 s later."

The first part of the question is: "How much higher or lower is the launch point relative to the point where the projectile hits the ground?
Express a launch point that is lower than the point where the projectile hits the ground as a negative number."

Which I already found out using y=(viy)(t)-(1/2)(g)(t^2)

Substituting the given numbers, I get -180 m, which would be the case if the landing point was below the launch point. Since the question wants a launch point lower than the landing point, the answer is 180 m.

The second part, where I'm having trouble: "To what maximum height above the launch point does the projectile rise?"

What I have,

x=(26)(cos(40))(t)
y=(26)(sin(40))(t)-(1/2)(g)(t^2)

I can't substitute 180 m or 8 s because those numbers apply to the landing point, and I'm looking for ymax so I end up with three unknowns in two equations, which I cannot solve. I don't know how to set this up so that I can solve it.

Thanks for any help!

Last edited: Oct 29, 2006
2. Oct 29, 2006

### rsk

To find maximum height, remember that the vertical component of velocity at the maximum height is zero.

Does this help?

3. Oct 29, 2006

### Hootenanny

Staff Emeritus
Hint: Try using the formulae;

$$v_{f}^{2}=v_{i}^{2}+2ax$$

Edit: Beaten to it.

4. Oct 31, 2006

### SA32

Whoops! Should post here to let you know I used your hints to figure it out, so thanks!