Question

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.

Answer 1

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.

Answer 2

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

Answer 3

this came from a test bank so the second answer listed is correct but wasn't an option (you need the negative sign on g), so you had to do what the third person did so that h isn't in the answer. But, h was given in the original question so it should be allowed in the answer and I like the 2nd answer better.

Answer 4

Yes