an object of 3 kg is dropped from 2m above the ground. it takes a time interval of tE to reach the ground of the earth. if it is dropped on the moon, how long does it take to reach the ground? the a dure to g on the moon is 1/6 of that on the earth.
h= 1/2 g t^2
t= sqrt(2h/g)
so, if g is 1/6, it will take sqrt(6) times longer.
To determine how long it takes for the object to reach the ground on the moon, we need to use the principle of free fall and consider the acceleration due to gravity on the moon.
On Earth:
Mass of the object (m) = 3 kg
Height from which it is dropped (h) = 2 m
Acceleration due to gravity on Earth (gE) = 9.8 m/s^2 (standard value)
Using the kinematic equation: h = (1/2)gt^2 + u*t, where u is the initial velocity (which is 0 in this case), we can solve for time (tE) it takes for the object to reach the ground on Earth.
Substituting the known values, we get:
2 = (1/2)(9.8)tE^2
Simplifying the equation, we have:
tE^2 = 2 / (1/2)(9.8)
tE^2 = 0.408
Taking the square root of both sides, we find:
tE = 0.639 s (approximately)
On the Moon:
Acceleration due to gravity on the moon (gM) = (1/6) * gE
= (1/6) * 9.8
= 1.6333 m/s^2 (approximately)
Using the same kinematic equation, but with the acceleration due to gravity on the moon (gM) and the same height (h = 2 m), we can solve for the time (tM) it takes for the object to reach the ground on the moon.
2 = (1/2)(1.6333)tM^2
tM^2 = 2 / (1/2)(1.6333)
tM^2 = 2.448
tM = 1.5636 s (approximately)
Therefore, it will take approximately 1.5636 seconds for the object to reach the ground on the moon.