If the faster stone takes 10 seconds to return to the ground, how long will it take the slower stone to return?

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1 Answer
Feb 13, 2018

"please review the details below."

Explanation:

"let elapsed time to peak point for faster stone be " t_1

3v:"velocity of faster stone"

t_1=(3v)/g

"let elapsed time to peak point for slower stone be " t_2

v:"velocity of slower stone"

t_2=(v)/g

"Let's find the ratio " t_1 / t_2

t_1/t_2=(3cancel(v))/cancel(g)* cancel( g)/cancel(v)

t_1/t_2=3

"Since the movement is symmetric, the time to reach "
"the maximum point is equal to half of the duration of the flight."

t_1=10/2=5 sec.

5/t_2=3

t_2=5/3 sec.

"flight time for slower stone "=2*5/3=10/3 sec.

H=v^2/(2g)" slower stone's maximum height"

x=(3v)^2/(2g)=(9v^2)/(2g)" faster stone's maximum height"

"let find the ratio " H/x

H/x=cancel(v^2)/(cancel(2g))*(cancel(2g))/(9cancel(v^2)

H/x=1/9

x=9H