# Help with projectile physics

1. Apr 12, 2004

### jennypear

projectile is fired w/an angle of 28.0degrees above the horizontal and from height 46.0m agove the ground. the projectile strikes ground w/a speed of 1.95xVo. Find Vo

I started using the eqution
(Vfy)^2=(Voy)^2+2*Ay*(delta Y)

Voy=Vosin(theta)=.47Vo
Vfy=1.954*Vosin(theta)=.92Vo

(.92Vo)^2=(.47Vo)^2+(2*-9.8*-46)
Vo=37.9

but my answer is wrong and I can't think of another way to answer the problem.....Any ideas?

2. Apr 12, 2004

### sridhar_n

What is Vo? Is it the velocity of projection, or, is it just a constant?

Last edited: Apr 12, 2004
3. Apr 12, 2004

There's no guarantee that the projectile hits the ground at the same angle as it was launched!

4. Apr 13, 2004

### sridhar_n

However, for this problem, I think we should assume that the projectile hits the ground at the same angle as it was launched....

Sridhar

5. Apr 13, 2004

Why should that be assumed?

6. Apr 13, 2004

### Chen

Not only are you not supposed to make that assumption, it is wrong! The projectile is travelling in a parabola.........

You need to use the conservation of energy equation:

$$0 = \Delta E_m = \Delta E_p + \Delta E_k$$
$$0 = mg(0 - h) + \frac{1}{2}m(v_f^2 - v_0^2)$$
$$2gh = (1.95v_0)^2 - v_0^2 = 2.8025v_0^2$$

The initial velocity is 17.94m/s (for g = 9.8m/s2).

Last edited: Apr 13, 2004
7. Apr 13, 2004

### jennypear

Thanks for the help!