Question

Two stones released at the same time with one thrown vertically upward from the ground with a velocity of 20m/s and the other dropped from a height collided with each other 2 seconds after their release, determine this
height (g = 9.8m/s)​

Answer 1

Answer:

20.4 m

Explanation:

displacement by the first stone

= ut+1/2 gt^2

= 40 - 9.8 {we take downwards negative}

= 30.2

displacement for thr second

= 1/2 gt^2

= -9.8 {downward negative}

h = displacement covered by the first stone + displacement covered by the second stone

h = 30.2 - 9.8

= 20.4 m