Posted by **James** on Friday, May 18, 2012 at 4:43pm.

What is the velocity at the midway point of a ball able to reach a height y when thrown with an initial velocity

v0?

(Assume the ball is thrown upward and that up is the positive direction. Use the following as necessary: y and g. )

Someone posted this question before and was given the answer

v= sqrt (v_o)^2 + 9.8y

This answer is apparently wrong.

- Physics -
**bobpursley**, Friday, May 18, 2012 at 5:07pm
Maybe it is a reading of the question. If the max height is y, midpoint is y/2

If that is so, then the 9.8 should be -9.8, and the equation is dead on correct.

- Physics -
**Elena**, Friday, May 18, 2012 at 5:52pm
The motion to the highest point (height “H”)

v = vₒ - g•t,

v=0, t = vₒ/g.

H = vₒ•t - g•t²/2 = vₒ• vₒ/g –(g/2) •(vₒ/g)² = vₒ²/2•g.

Velocity at the midpoint (height “h“)

v =vₒ - g•t1,

t1 =(vₒ- v)/g.

h = vₒ•t1 - g•t1²/2 =

=vₒ•(vₒ- v)/g - g•(vₒ- v)²/2•g² =

= [2•vₒ•(vₒ- v) -(vₒ- v)²]/ 2•g =

={2vₒ² - 2vₒ•v - vₒ² +2 vₒ•v - v²}/2•g =

= (vₒ² - v²)/2•g.

h=H/2 = vₒ²/2•2•g.

(vₒ² - v²)/2•g = vₒ²/2•2•g.

vₒ² = 2•v²,

v =vₒ/√2 = vₒ/1.41 =0.707•vₒ

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