Question

a stone dropped from top of tower of height 490m.simultaneously another stone is projected up vertically with a velocity of 100m/s from the ground..the stone meet after time of​

Answer 1

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Answer -

For the first stone, we have:

h(t)=200-g/2 t²

where h(t) is the height at time t secon-ds.

Then, the thrown stone would be:

h(t)=40t-g/2t²

So, they meet at:

40t-g/2t²=200-g/2t²

40t=200

t=5 seconds

h(5)=200–4.9(25)

=77.5 m above the ground……….