Question

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? Assume free fall

Answer 1

Answer:

$t_{slow} = 3.33\ s$

Explanation:

given,

stone 1 velocity,u' = v

stone 2 velocity, u = 3 v

time of the faster stone = 10 s

time of the slower stone = ?

using equation of motion

$s = ut + \dfrac{1}{2}gt^2_{fast}$

displacement , s = 0

$0 = ut - \dfrac{1}{2}gt_{fast}^2$

$t_{fast} = \dfrac{2u}{g}$.........(1)

$t_{fast} = 3 \dfrac{2v}{g}$

$\dfrac{t_{fast}}{3} =\dfrac{2v}{g}$.......(2)

time taken by the slower ball

$t_{slow} = \dfrac{2u'}{g}$

$t_{slow} = \dfrac{2v}{g}$

from equation (2)

$t_{slow} = \dfrac{t_{fast}}{3}$

$t_{slow} = \dfrac{10}{3}$

$t_{slow} = 3.33\ s$

time taken by the slower stone is equal to 3.33 s