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

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The calculation of initial velocity (Vo) for a projectile launched at an angle of 28.0 degrees from a height of 46.0 meters is determined using the conservation of energy principle. The correct initial velocity is found to be 17.94 m/s, derived from the equation 2gh = (1.95Vo)^2 - Vo^2, where g is 9.8 m/s². The initial assumption that the projectile hits the ground at the same angle as it was launched is incorrect, as projectiles follow a parabolic trajectory. This discussion clarifies the importance of using energy conservation rather than kinematic equations alone for accurate results.

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jennypear
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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?
 
Last edited:
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).
 
Last edited:
Thanks for the help!
 

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