Question

a stone is dropped from a height of 49m and simultaneusly another ball is thrown upwards from the ground with a speed of 40m/s .When and where do the stones meet​

Answer 1

You mentioned this speed of 40m/s because this is the speed that the stone which is thrown from a height of 49m will reach when it touches the ground.

After a time t, the height reached by the stone thrown from the ground will be:

40*t-1/2*g*t²

After a time t the height of the stone thrown from 49m will be:

49-1/2*g*t²

The two stones will meet when their height is the same, so at a time t which is such that:

40*t-1/2*g*t²=49-1/2*g*t²

So, by eliminating the same member on the two sides of this equality:

40*t=49

which gives: t=49/40=1.225 second.

And it corresponds to a height of 41.64 meters.

hope it helps...