- #1
Ace.
- 52
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[itex][/itex]
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.
v22 = v12 + 2ad ?
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?
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?