Height when a ball is thrown vertically at half its velocity

  • Thread starter Ace.
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  • #1
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[itex][/itex]

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 = [itex]\frac{v^{2}_{2} - v^{1}_{2}}{2g}[/itex]
d = [itex]\frac{0 m/s - (10 m/s)^{2}}{2(-9.8 m/s^{2}}[/itex]
d = 5.1

at 5 m/s:


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

[itex]\frac{1.28}{5.1}[/itex] = 0.25

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

Answers and Replies

  • #2
Dick
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[itex][/itex]

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 = [itex]\frac{v^{2}_{2} - v^{1}_{2}}{2g}[/itex]
d = [itex]\frac{0 m/s - (10 m/s)^{2}}{2(-9.8 m/s^{2}}[/itex]
d = 5.1

at 5 m/s:


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

[itex]\frac{1.28}{5.1}[/itex] = 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.
 
  • #3
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I didn't fully understand what you mean
 
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
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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:
  • #5
Dick
Science Advisor
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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.
 

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