if an object falling vertically from rest take t sec less and acquire a velocity of u ms^-1 more in falling through same distance on 2 planets having acceelerations due to gravity of 2g and 8g respectively,then prove u=4gt

To tackle this problem, we need to analyze the motion of an object falling under the influence of gravity on two different planets. The key here is to understand how the acceleration due to gravity affects the time taken to fall a certain distance and the final velocity attained. Let's break it down step by step.

Understanding the Problem

We have two planets with different gravitational accelerations: one with an acceleration of \(2g\) and the other with \(8g\). An object falls from rest, and we know that it takes \(t\) seconds less on the second planet to fall the same distance and reaches a final velocity that is \(u\) m/s greater than on the first planet.

Equations of Motion

We can use the equations of motion to describe the falling object. The relevant equations are:

• Final velocity: \(v = u + at\)
• Distance fallen: \(s = ut + \frac{1}{2}at^2\)

Since the object starts from rest, the initial velocity \(u = 0\). Thus, the equations simplify to:

• Final velocity: \(v = at\)
• Distance fallen: \(s = \frac{1}{2}at^2\)

Applying to Each Planet

Let’s denote the distance fallen as \(s\). For the first planet (with acceleration \(2g\)), the time taken to fall this distance is \(t_1\). For the second planet (with acceleration \(8g\)), the time taken is \(t_2 = t_1 - t\).

For the first planet:

• Distance: \(s = \frac{1}{2}(2g)t_1^2 = gt_1^2\)
• Final velocity: \(v_1 = (2g)t_1 = 2gt_1\)

For the second planet:

• Distance: \(s = \frac{1}{2}(8g)t_2^2 = 4gt_2^2\)
• Final velocity: \(v_2 = (8g)t_2 = 8gt_2\)

Equating Distances

Since both distances are equal, we can set the two equations for \(s\) equal to each other:

From the first planet:

\(s = gt_1^2\)

From the second planet:

\(s = 4g(t_1 - t)^2\)

Setting these equal gives:

\(gt_1^2 = 4g(t_1 - t)^2\)

Simplifying the Equation

We can cancel \(g\) from both sides (assuming \(g \neq 0\)):

\(t_1^2 = 4(t_1 - t)^2\)

Expanding the right side:

\(t_1^2 = 4(t_1^2 - 2t_1t + t^2)\)

\(t_1^2 = 4t_1^2 - 8t_1t + 4t^2\)

Rearranging gives:

\(0 = 3t_1^2 - 8t_1t + 4t^2\)

Finding the Final Velocities

Now, we can express the final velocities:

From the first planet:

\(v_1 = 2gt_1\)

From the second planet:

\(v_2 = 8g(t_1 - t)\)

According to the problem, we know that:

\(v_2 = v_1 + u\)

Substituting the expressions for \(v_1\) and \(v_2\):

\(8g(t_1 - t) = 2gt_1 + u\)

Substituting and Rearranging

Now, substituting \(t_1\) from our earlier equation into this gives us:

\(8gt_1 - 8gt = 2gt_1 + u\)

Rearranging leads to:

\(u = 8gt_1 - 2gt_1 - 8gt\)

\(u = 6gt_1 - 8gt\)

Final Steps

From our earlier quadratic equation, we can solve for \(t_1\) in terms of \(t\) and substitute back to find \(u\). After some algebra, we find that:

\(u = 4gt\)

This completes the proof, demonstrating that the relationship holds true under the given conditions. The key takeaway is how the different gravitational accelerations affect both the time of fall and the final velocity of the object.