Question

A stone is thrown vertically upwards with a velocity of 19.6 m/s. After 2 second, another stone

Is thrown upwards with a velocity of 9.8 m/s. When and where will these stones code
(take g = 9.8 m/s)​

Answer 1

speed of first stone, u1 = 19.6 m/s

speed of 2nd stone , u2 = 9.8 m/s

a/c to question,

2nd stone is thrown after 2sec. let's see after 2sec velocity of 1st stone , v = u + at

here , u = u1 = 19.6 m/s and a = -9.8m/s²

so, v = 19.6 + (-9.8)(2) = 0

hence, after 2sec 1st stone start to fall downward.

height of 1st stone at that time, s = 19.6 × 2 - 1/2 × 9.8 × 2² = 19.6 m.

hence, seperation between 1st and 2nd stone is 19.6m

now, time taken by stones to collapse , t = seperation between stone/relative velocity

= 19.6m/(19.6m/s - 9.8m/s)

= 2sec

and height of stones from the ground where they collapse , h = distance covered by 1st stone in 2sec

= 9.8 × 2 - 1/2 × 9.8 × 2²

= 19.6 - 19.6 = 0

hence, they collapse on the ground.