Physics
posted by James on .
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.

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. 
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