Respuesta :
Answer:
[tex]\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )[/tex]
Explanation:
Given:
height above which the rock is thrown up, [tex]\Delta h=50\ m[/tex]
initial velocity of projection, [tex]u=20\ m.s^{-1}[/tex]
let the gravity on the other planet be g'
The time taken by the rock to reach the top height on the exoplanet:
[tex]v=u+g'.t'[/tex]
where:
[tex]v=[/tex] final velocity at the top height = 0 [tex]m.s^{-1}[/tex]
[tex]0=20-g'.t'[/tex] (-ve sign to indicate that acceleration acts opposite to the velocity)
[tex]t'=\frac{20}{g'}\ s[/tex]
The time taken by the rock to reach the top height on the earth:
[tex]v=u+g.t[/tex]
[tex]0=20-g.t[/tex]
[tex]t=\frac{20}{g} \ s[/tex]
Height reached by the rock above the point of throwing on the exoplanet:
[tex]v^2=u^2+2g'.h'[/tex]
where:
[tex]v=[/tex] final velocity at the top height = 0 [tex]m.s^{-1}[/tex]
[tex]0^2=20^2-2\times g'.h'[/tex]
[tex]h'=\frac{200}{g'}\ m[/tex]
Height reached by the rock above the point of throwing on the earth:
[tex]v^2=u^2+2g.h[/tex]
[tex]0^2=20^2-2g.h[/tex]
[tex]h=\frac{200}{g}\ m[/tex]
The time taken by the rock to fall from the highest point to the ground on the exoplanet:
[tex](50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2[/tex] (during falling it falls below the cliff)
here:
[tex]u=[/tex] initial velocity= 0 [tex]m.s^{-1}[/tex]
[tex]\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2[/tex]
[tex]t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}[/tex]
[tex]t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }[/tex]
Similarly on earth:
[tex]t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }[/tex]
Now the required time difference:
[tex]\Delta t=(t'+t_f')-(t+t_f)[/tex]
[tex]\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )[/tex]
