Question

A ball is thrown directly upwards at 25 ms and the same time another ball is released from height of loom. At what height and time will the balls meets?​

Answer 1

Step-by-step explanation:

When the balls meets, the time should be the same, let say t.

For the first ball (released from a loom):

`\bar{h}=(1)/(2)gt^(2)` (Free fall motion) (
`\bar{h}` = the distance travelled by the first ball, measured from the loom)

So, we can find that
`\bar{h}=(1)/(2)(10t^(2))=5t^(2)`

For the second ball (which is thrown directly upwards):

`y=v_(o)t-(1)/(2)gt^(2) = 25t-5t^(2)`

Because the height of the loom is h :

`h=\bar{h}+y`

`h=25t`

So, the time when two balls meets is :
`t=(h)/(5)` (In this question, the height of the loom h is not declared). When the height of the loom is known, you can calculate the time and the value of
`\bar{h}` or y