a stone is thrown vertically upward with velocity 19.6 m/s . after 2second another stone is thrown upward with velocity 9.8m/s. when where will these stone collide?(g=9.8m/s²)

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Solution

Initial speed of the first stone = 19.6 m/s
Upward distance covered by this stone
$v^{2} - u^{2} = 2 a s v = 0 u = 19.8 m / s a = - g w e g e t s = 19.6 m$

because at top most point final velocity will be zero.
time taken to cover this height is
$s = u t + 1 / 2 a t^{2} s = 19.6 m u = 19.6 m / s a = - g w e g e t t = 2 s$

so it will take same time to reach ground
total time taken by first stone to reach ground = 2+2 = 4s

this means when first stone is at the top most point then second stone is projected upward.
similarly for the second stone upward distance travelled will be 4.9 m and time taken to reach at the top most point will ne 1 second after the time of projection.
it will also take same time to reach ground so total time of whole journey of the second stone will be 1+1= 2 s after the projection time.
As stone was projected after 2 s then net time is 2+2 = s.

Therefore after the second both the stones will come to the ground and they will collide at the ground.

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a stone is thrown vertically upward with velocity 19.6 m/s . after 2second another stone is thrown upward with velocity 9.8m/s. when where will these stone collide?(g=9.8m/s2)

a stone is thrown vertically upward with velocity 19.6 m/s . after 2second another stone is thrown upward with velocity 9.8m/s. when where will these stone collide?(g=9.8m/s²)