Finding initial vertical velocity given acceleration

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SUMMARY

The discussion centers on calculating the initial vertical velocity of a shot put when released after a constant upward acceleration of 45.4 m/s² over a height of 65.0 cm. The user initially misapplies the kinematic equations, assuming the final velocity is zero at the release point. The correct approach involves using the equation V² = initial V² + 2a(x - initial x) to find the initial velocity, which results in a value of 7.68 m/s. The user is advised to ensure the problem statement is accurately represented.

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freshyy
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Homework Statement


Sam heaves a shot with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 45.4 m/s^2 for a height 65.0 cm . He releases it at height 2.16 m above the ground. You may ignore air resistance.

Q-What is the speed of the shot when he releases it?

Homework Equations

: [/B]
Equations I believe I can use are
V=initial V +at or
V^2=initial V^2 +2a(x-initial x)

The Attempt at a Solution


Since the final velocity would be zero when it is at the highest point i got
0=initial V^2 +2(45.4m/s^2)(-.65m)
I would have V ^2=59.02
then V=7.68m/s
but this says it is wrong when i check my answer.
Please help
 
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freshyy said:
Since the final velocity would be zero when it is at the highest point i got
The final velocity is the velocity when he releases it--that's not zero. It starts at zero.

Nonetheless, the equation is correct and so is your answer. Are you sure you are posting the exact problem?
 

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