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

Answer:

Explanation:

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

For the first ball (released from a loom):

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

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

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

$y=v_{o}t-\frac{1}{2}gt^{2} = 25t-5t^{2}$

Because the height of the loom is h :

$h=\bar{h}+y$

$h=25t$

So, the time when two balls meets is : $t=\frac{h}{5}$ (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