Question:

A stone is thrown vertically upwards with a velocity of 19.6 ms^-1 . After 2 seconds, another stone is thrown upwards with a velocity of 9.8 ms^-1 .When and where will these stones collide?

Solution:

First Stone:

Time taken by the first stone to reach its maximum height when thrown with an initial speed of 19.6 ms^-1

$${t=\frac{u}{g}}$$

$${t=\frac{19.6}{9.8}}$$

$${t=2 \ sec}$$

Total time to hit the ground is 4 sec.

Second stone:

Time taken by second stone to reach its maximum height when thrown with an initial speed of 9.8 ms^-1

$${t=\frac{9.8}{9.8}}$$

$${t=1 \ sec}$$

Total time to hit the ground is 2 sec.

After 4 seconds:

The first stone will hit the ground.The second stone will also hit the ground (because it was thrown after 2 seconds)

Therefore both the stones will collide at the ground.