projectile

[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?

Thanks for your time!

[sridhar_n]

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

[cookiemonster]

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

cookiemonster

[sridhar_n]

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

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


Sridhar

[cookiemonster]

Why should that be assumed?

cookiemonster

[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).

[jennypear]

Thanks for the help!