a body of mass m falls from a height h to ground if coefficient of restitution is e then calculate the distance travelled by body before it comes to rest
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to solve this let's find the relation between the height raised of the ball when it falls from a height 'h'.
let 'v' be the velocity of the ball when it hits the ground.
a = g
so v = at (let 't' be the time taken to hit the ground)
=> t = v/a
=> h = 0 x t + (1/2) a t^2 (t^2 -> t square)
so h = (v x v) / 2a
then it rebonds with 'ev' velocity.
so the height it reaches is h' = (ev) t1 - (1/2) a t1^2 (t1^2 -> t1 square)
t1 = ev/a (v = u-at and v = 0 and u = ev here)
so by substituting t1 in h' we get
h' = (e square) x(v square) / 2a = (e square) x h
so distance travelled = h + (e^2) h + (e^4) h + (e^6) h + ......... which is in gp
so total distance = (h / (1-(e square)))
let 'v' be the velocity of the ball when it hits the ground.
a = g
so v = at (let 't' be the time taken to hit the ground)
=> t = v/a
=> h = 0 x t + (1/2) a t^2 (t^2 -> t square)
so h = (v x v) / 2a
then it rebonds with 'ev' velocity.
so the height it reaches is h' = (ev) t1 - (1/2) a t1^2 (t1^2 -> t1 square)
t1 = ev/a (v = u-at and v = 0 and u = ev here)
so by substituting t1 in h' we get
h' = (e square) x(v square) / 2a = (e square) x h
so distance travelled = h + (e^2) h + (e^4) h + (e^6) h + ......... which is in gp
so total distance = (h / (1-(e square)))
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