Question

A stone is thrown vertically upward at a speed of 20.10 m/s at time t=0. A second stone is thrown upward with the same speed 1.510 seconds later.

At what time are the two stones at the same height?

At what height do the two stones pass each other?

What is the downward speed of the first stone as they pass each other?

Answer 1

Given is:-

Initial speed of both the stones is u = 20.10 m/s

Now for first stone

    eq-1

for second stone

eq-2

now equating both equations we get

which gives us

Part - b

By plugging the value in eq-1 we get

which gives us

This is the height where tow stones pass each other.

part - c

Negative signs shown that the velocity is downwards.