# Height when a ball is thrown vertically at half its velocity



## 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 Helper


## 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.

I didn't fully understand what you mean

NascentOxygen
Staff Emeritus
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?

Last edited:
Dick