The acceleration due to gravity on planet X is

one fifth that on the surface of the earth.
If it takes 4.6 s for an object to fall a certain
distance from rest on earth, how long would
it take to fall the same distance on planet X?
Answer in units of s

The acceleration due to gravity on planet X is one fifth that on the surface of the earth. Let's denote the acceleration due to gravity on Earth as 'g' and on planet X as 'g_x'.

Given: g_x = (1/5)g

We can use the equation of motion for free fall:

distance = (initial velocity * time) + (1/2 * acceleration * time^2)

For both cases (Earth and planet X), the distance is the same. Let's assume the distance as 'd' and the time taken on planet X as 't_x'.

For Earth:
distance = d
initial velocity = 0 (object starts from rest)
acceleration = g
time = 4.6 s

Using the equation of motion, we have:
d = (0 * 4.6) + (1/2 * g * 4.6^2)
d = 2.3 * g

For planet X:
distance = d
initial velocity = 0 (object starts from rest)
acceleration = g_x
time = t_x (to be determined)

Using the equation of motion, we have:
d = (0 * t_x) + (1/2 * g_x * t_x^2)
d = 0.5 * g_x * t_x^2

We can substitute the value of g_x from the given information:
d = 0.5 * (1/5g) * t_x^2
d = (1/10g) * t_x^2

Since the distance is the same on both Earth and planet X (d = 2.3 * g = (1/10g) * t_x^2), we can equate these two expressions:

2.3 * g = (1/10g) * t_x^2

Simplifying, we get:
23g^2 = t_x^2

Taking the square root of both sides, we have:
t_x = √(23g^2)

Substituting the value of g as 9.8 m/s^2 (acceleration due to gravity on Earth), we get:
t_x = √(23 * (9.8)^2)
t_x ≈ √(23 * 96.04)
t_x ≈ √(2209.92)
t_x ≈ 47.02 s

Therefore, it would take approximately 47.02 seconds for the object to fall the same distance on planet X.

To find the time it would take for an object to fall the same distance on planet X, we can use the relationship between time, distance, and acceleration due to gravity.

Let's assume the time it takes for the object to fall the same distance on planet X is t seconds.

Given that the acceleration due to gravity on planet X is one-fifth that on the surface of the earth, we can say that the acceleration on planet X (aX) is one-fifth of the acceleration on Earth (aE). Mathematically, we can write this as:

aX = (1/5) * aE

Now, we know that the distance fallen is the same on both Earth and planet X. Let's call this distance d.

Using the equation for distance fallen with constant acceleration:

d = (1/2) * a * t^2

On Earth:
dE = (1/2) * aE * tE^2

On planet X:
dX = (1/2) * aX * t^2

Since we are considering the same distance, we can equate the equations for dE and dX:

(1/2) * aE * tE^2 = (1/2) * aX * t^2

We can now substitute the given values into the equation.

Given:
aX = (1/5) * aE
tE = 4.6 seconds

Substituting these values into the equation, we get:

(1/2) * aE * (4.6)^2 = (1/2) * ((1/5) * aE) * t^2

Simplifying the equation, we get:

(4.6)^2 = (1/5) * t^2

Now, we can solve for t:

t^2 = (5 * (4.6)^2)
t^2 = 105.64
t = sqrt(105.64)
t ≈ 10.28 seconds

Therefore, it would take approximately 10.28 seconds for an object to fall the same distance on planet X.