a stone is droppedfrom height of 100m fromgroundand 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
Answers
Answered by
1
They will meet instinctively. Here's how.
Lets take distance covered by the stone thrown upwards to meet the other as x. Then the other stone covers a distance of (100-x)m.
Let the stone dropped down be s1. So the stone thrown up is s2.
s1 => d = 100-x
g = 9.8 m/s²
u = 0
We know, s= ut + 1/2at²
Putting values,
100-x = 4.9t² ....(1)
s2 => d = x
g = -9.8 m/s²
u = 25 m/s
We know, s= ut + 1/2at²
Putting values,
x = 25t -4.9t² ....(2)
Adding (1) and (2),
100-x +x = 4.9t² + 25t - 4.9t²
t = 4 secs.
Puuting t = 4 in (1),
100-x = 4.9t²
100-x =19.6
x = 80.4 m
So, they meet at 80.4 m from ground after 4 seconds.
Similar questions