Height when a ball is thrown vertically at half its velocity

AI Thread Summary
A baseball thrown vertically reaches a maximum height, h, with an initial velocity, v. To find the height at which the baseball has half its original velocity, it's important to recognize that the calculations should be based on the distance from the ground, not from the peak of the trajectory. The correct approach involves determining the height where the velocity is v/2, rather than calculating distances from the apex where velocity is zero. The calculations provided initially misinterpret the reference level, leading to confusion about the resulting height. Ultimately, the correct height at half the original velocity is 0.75 of the maximum height reached.
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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?
 
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Ace. said:

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
 
Ace. said:
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:
Ace. said:
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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