What is the velocity at the midway point of projectile motion?

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SUMMARY

The velocity at the midway point of projectile motion for a ball thrown with an initial velocity \( v_0 \) can be determined using the equation \( v = \sqrt{\frac{v_0^2}{2}} \). The discussion clarifies that at the maximum height, the acceleration \( a \) can be expressed as \( a_y = -\frac{v_0^2}{2} \). This relationship is derived from the kinematic equation \( v^2 = v_0^2 + 2a_y \), where \( y \) represents the maximum height reached by the projectile. The simplification process is crucial for arriving at the correct expression for velocity at the midpoint.

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rasen58
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Homework Statement


The velocity at the midway point of a ball able to reach a height y when thrown with a velocity v0 at the origin is:

Homework Equations


v2=v02+2ay

The Attempt at a Solution


I made y = 1/2 y and the simplified the equation to v=sqrt(v02+ay)
But that is wrong because the answer is sqrt(v02/2)
Why?
 
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See if you can continue to simplify your result. In particular, can you find a way to express ay in terms of v0?
 
a = (v-v0)/t
y = v0t+1/2 at2
Substituting those don't seem to help enough
 
In your first post, you stated the following useful equation: v2 = vo2 + 2ay.

In your problem, y represents the height to which the ball will reach (i.e., the maximum height). When the ball reaches maximum height, what can you say about the value of v on the left side of the equation?
 
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Oh wow, then you get ay = - v02/2
and then if you plug that back into my first equation, you get the answer.
Wouldn't have thought of that. Thanks!
 
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Can you please explain more ?
 

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