Height when a ball is thrown vertically at half its velocity

[Ace.]

Homework Statement

A baseball is thrown vertically into the air with a velocity, v, and reaches a maximum height, h. At what height was the baseball with one-half its original velocity? Assume no air resistance.

Homework Equations

v₂² = v₁² + 2ad ?

The Attempt at a Solution

let 10m/s be the original velocity
d = \frac{v^{2}_{2} - v^{1}_{2}}{2g}
d = \frac{0 m/s - (10 m/s)^{2}}{2(-9.8 m/s^{2}}
d = 5.1

at 5 m/s:


d = \frac{0 m/s - (5 m/s)^{2}}{2(-9.8 m/s^{2}}
d = 1.28

\frac{1.28}{5.1} = 0.25

but the answer is 0.75 of the original height. How to solve this?

 

[Dick]

Homework Statement

A baseball is thrown vertically into the air with a velocity, v, and reaches a maximum height, h. At what height was the baseball with one-half its original velocity? Assume no air resistance.

Homework Equations

v₂² = v₁² + 2ad ?

The Attempt at a Solution

let 10m/s be the original velocity
d = \frac{v^{2}_{2} - v^{1}_{2}}{2g}
d = \frac{0 m/s - (10 m/s)^{2}}{2(-9.8 m/s^{2}}
d = 5.1

at 5 m/s:


d = \frac{0 m/s - (5 m/s)^{2}}{2(-9.8 m/s^{2}}
d = 1.28

\frac{1.28}{5.1} = 0.25

but the answer is 0.75 of the original height. How to solve this?

The d you are calculating is the distance between the point where the velocity v=0 and the point where the velocity is v/2. The point where v=0 is at the TOP of your trajectory.

 

[Ace.]

I didn't fully understand what you mean

 

[NascentOxygen]

I didn't fully understand what you mean

Ace, in the second part are not using the correct reference level.

The question you should be answering is: if a ball is thrown vertically upwards, at what height above the ground has its velocity dropped back to half of what it initially had?

 

[Dick]

I didn't fully understand what you mean

You are calculating distances from the point where v=0. So the second distance you calculated isn't the distance from the ground. It's distance from the point where v=0. v=0 is at the top of the trajectory.