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? (b) if the slower stone reaches a maximum height of h, how high (in terms of h) will the faster stone go? assume free fall.

t₁ = 3.33s, h = 9 H

Explanation:

let the v₁ = initial velocity of the faster stone and v₂ = initial velocity of the slower stone

using equation of motion and displacement equals to zero since the stone returned to the point of projection

y - y₀ =  v₁ t - 1/2gt²

- v₁ t  = - 1/2gt²

2v₁ t / t² = g

g = 2v₁ / t

repeat the same produce for the slower stone where the time =   t₁

y-y₀ =  v₂t₁ - 1/2 gt₁²

- v₂t₁ = - 1/2 gt₁²

t₁ = 2v₂ / g = 2v₂ / (2v₁ / t) = (2v₂ / 2v₁) × t

and

v₁  = 3v₂

t₁ = (2v₂ / 2v₁) × t  = (v₂ / 3v₂) × 10 = 3.33 s

b) using the equation of motion

vf₂² = v₂² - 2gH

since the body stop momentarily at maximum height

- v₂²  = - 2gH

v₂²  / 2H = g

repeating the same procedure for the faster stone

vf₁² = v₁² - 2gh

- v₁² =  - 2gh

v₁²/ 2g = h

substitute for g

h = v₁² / 2(v₂²  / 2H ) = (v₁² / v₂²) × H = (3v₂)² / (v₂² ) × H = 9H

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat