# Projectile Motion diver leap

1. Nov 25, 2008

### ahrog

1. The problem statement, all variables and given/known data
A diver leaps from the top of a tower with an initial horizontal velocity of 5.0 m/s and an upward velocity of 2.0 m/s. Use a vector diagram to indicate the diver's horizontal and vertical velocity components 1.0s after he leaves the tower.

2. Relevant equations
x=vxt
y= vyt + 1/2gt2
maybe v=d/t

3. The attempt at a solution

I'm not exactly sure how to go about this problem. Do I just use the x and y formula, and that is all?
Vx= x/t
=5/1
=5

Vy= (y-1/2gt2)/t
=(2-1/2(9.8)(1)2)/1
=-2.9

Would those be right?

2. Nov 26, 2008

### ak1948

Not quite right. The best equation to use here is v2 - v1 = a(t2 - t1). So for vx, a =0 and v2 = v1 = 5. (as you said). For vy, a = -9.8 and v2 = 2 - 9.8 = -7.8.

3. Nov 26, 2008

### ahrog

Really? They haven't mentioned that formula in what I've done so far...

4. Nov 27, 2008

### Mentallic

I had never witnessed such a formula either till it clicked - it's just a slight modification to the $$v=u+at$$ formula.
where:
v = final velocity
u = initial velocity
a = acceleration due to gravity
t = time

And yes, this is the best formula to use in this case. Horizontal velocity will be kept constant at 5ms-1 since we are assuming no resistances, so you already know what that vector is going to be after 1 second. Now you just need to calculate the vertical vector with this equation, which ak1948 already gave as -7.8ms-1.