Why Is My Calculation of Initial Velocity Incorrect in Projectile Motion?

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The discussion revolves around calculating the initial velocity (Vo) of a projectile launched at a 28-degree angle from a height of 46 meters. The initial calculations using the vertical motion equation yielded an incorrect Vo of 37.9 m/s. Participants clarify that the assumption of the projectile hitting the ground at the same angle as it was launched is incorrect, emphasizing the parabolic trajectory of the motion. Instead, they suggest applying the conservation of energy principle, leading to a corrected initial velocity of 17.94 m/s. The conversation highlights the importance of accurately applying physics principles in projectile motion problems.
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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!
 
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What is Vo? Is it the velocity of projection, or, is it just a constant?
 
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There's no guarantee that the projectile hits the ground at the same angle as it was launched!

cookiemonster
 
cookiemonster said:
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
 
Why should that be assumed?

cookiemonster
 
Not only are you not supposed to make that assumption, it is wrong! The projectile is traveling 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).
 
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Thanks for the help!
 
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