Question

a stone is droppedfrom height of 100m from ground and at the sametime another stone isthrownup withvelocity of 25m/s.calculate when and where both stones meet? also had the distancetravelled by bothstones then meet?

Answer 1

Hey there !!!!!

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Stone is dropped from a height of 100m from ground and simultaneously a stone is projected upwards from the ground

Let the distance from ground at which these stones meet = p

So,

Distance traveled by stone projected upwards when two stones meet = p

Distance traveled by stone dropped from a height of 100m when these stones meet = 100-p

Stone dropped from tower:

distance traveled = 100-p

          g = 9.8 m/s²

  initial velocity(u)=0

We know, s= ut + gt²/2

Putting values,

100-p =4.9t²-----Equation 1

Stone projected upwards:

 distance traveled = p

          g = -9.8 m/s².

          u = 25 m/s

We know, s= ut + 1/2gt².

Putting values,

p = 25t -4.9t²---Equation 2

Adding (1) and (2),

100-p+p = 4.9t²+ 25t - 4.9t².

100=25t

t=4 seconds

Putting t = 4 in (1),

100-p = 4.9t².

100-p =4.9*16

 100-4.9*16=p

100-78.4=21.6 m

So,two stones meet at height of 21.6m from ground.

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Hope this helped you............... 

 

Answer 2

Answer:

because of your lazyness