a ball rolls off a table and hits the floor at 5m/s. what is the height of the table.....
PLEASE FAST......
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Answered by
1
Dear Student,
When a ball rolls off a table, the time it takes to hit the floor is a function of the height of the table only and it does not depend on the speed.
T =
![\sqrt{2h \gamma g \sqrt{2h \gamma g](https://tex.z-dn.net/?f=+%5Csqrt%7B2h+%5Cgamma+g)
When a ball rolls off a table, the time it takes to hit the floor is a function of the height of the table only and it does not depend on the speed.
T =
souvikmondal78pcblkn:
Wring
Answered by
16
here velocity of the ball when it hit the ground
v = 5 m/s, let height be h
using third equation of kinematics
![{v}^{2} - {u}^{2} = 2gh {v}^{2} - {u}^{2} = 2gh](https://tex.z-dn.net/?f=+%7Bv%7D%5E%7B2%7D++-++%7Bu%7D%5E%7B2%7D++%3D+2gh)
![h \: = \frac{ {v}^{2} - {u}^{2} }{2g} h \: = \frac{ {v}^{2} - {u}^{2} }{2g}](https://tex.z-dn.net/?f=h+%5C%3A++%3D++%5Cfrac%7B+%7Bv%7D%5E%7B2%7D++-++%7Bu%7D%5E%7B2%7D+%7D%7B2g%7D+)
![h \: = \frac{ {(5m { s}^{ - 1}) }^{2} - {0}^{2} }{2 \times 9.8m {s}^{ - 2} } = 1.28 m h \: = \frac{ {(5m { s}^{ - 1}) }^{2} - {0}^{2} }{2 \times 9.8m {s}^{ - 2} } = 1.28 m](https://tex.z-dn.net/?f=h+%5C%3A++%3D++%5Cfrac%7B++%7B%285m+%7B+s%7D%5E%7B+-+1%7D%29+%7D%5E%7B2%7D++-++%7B0%7D%5E%7B2%7D+%7D%7B2+%5Ctimes+9.8m+%7Bs%7D%5E%7B+-+2%7D+%7D++%3D+1.28+m)
v = 5 m/s, let height be h
using third equation of kinematics
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