From Equations of motion

The height of a jump is inversely proportional to the acceleration due to gravity. Let \( h_A \) be the height of the jump on planet A, and \( h_B \) be the height of the jump on planet B. Also, let \( g_A \) and \( g_B \) represent the acceleration due to gravity on planets A and B, respectively.

Given:
- \( g_A = 9g_B \)
- \( h_A = 2 \, \text{m} \)

The ratio of heights is:

\[
\frac{h_B}{h_A} = \frac{g_A}{g_B} = 9
\]

Thus, the height on planet B is:

\[
h_B = 9 \times h_A = 9 \times 2 = 18 \, \text{m}
\]

So, the height of the jump on planet B is \( 18 \, \text{m} \).

Let the initial velocity of the ball be \( u \), and let \( v \) be the velocity of the ball at a distance \( h \) above the tower. Using the equation of motion:

\[
v^2 = u^2 - 2gh
\]

At a distance \( h \) below the tower, the velocity is doubled, so:

\[
(2v)^2 = u^2 + 2gh
\]

Simplifying these:

\[
4v^2 = u^2 + 2gh
\]

Substitute \( v^2 = u^2 - 2gh \) from the first equation:

\[
4(u^2 - 2gh) = u^2 + 2gh
\]

Expanding and solving:

\[
4u^2 - 8gh = u^2 + 2gh
\]

\[
3u^2 = 10gh
\]

\[
u^2 = \frac{10gh}{3}
\]

The maximum height \( H \) from the top of the tower is given by:

\[
H = \frac{u^2}{2g} = \frac{10gh}{6g} = \frac{5h}{3}
\]

Thus, the maximum height attained by the ball from the top of the tower is \( \frac{5h}{3} \).

From Galileo's ratio of odd number distances travelled in consecutive seconds are in the order of 1:3:5:7... Here distance travelled in 3rd second is 5/9th of total distance total time of flight is 3 second.

\[ s= \frac{1}{2}gt^{2}= \frac{1}{2}\times 10\times 3^{2}= 45 m \]

From Galileo's ratio of odd numbers distances travelled in consecutive seconds are in the order of 1:3:5:7

As distance in last second is is 7x. total distance = x+3x+5x+7x= 16x = 16×5= 80 m 

\[ \frac{1}{2}gt^{2} - \frac{1}{2}g(t-1)^{2} = 20 \]

solving we get, t =2.5 sec.

\[ h= \frac{1}{2}g(2.5)^{2}= 31.25 m \]

1. Time taken by each ball:
- First ball: \( t_1 = 5 \, \text{s} \)
- Second ball: \( t_2 = 3 \, \text{s} \) (dropped 2 s later)

2. Heights:
- Height of the first ball:
\[
h_1 = \frac{1}{2} g t_1^2 = \frac{1}{2} \cdot 10 \cdot (5)^2 = 125 \, \text{m}
\]
- Height of the second ball:
\[
h_2 = \frac{1}{2} g t_2^2 = \frac{1}{2} \cdot 10 \cdot (3)^2 = 45 \, \text{m}
\]

3. Difference in heights:
\[
\Delta h = h_1 - h_2 = 125 - 45 = 80 \, \text{m}
\]

 The difference in initial heights is 80 meters.

\[ 0.36 H= \frac{1}{2}g(t-1)^{2} \]

\[ H= \frac{1}{2}g(t)^{2} \]

Dividing we get

\[ 0.36 = \frac{(t-1)^{2}}{t^{2}} \]

\[ 0.6 = \frac{(t-1)}{t} \]

0.4t =1

t= 2.5 sec

S= (1/2) ×10 ×(2.5)²= 31.25 m

Concept: Equations of motion under gravity.
Formula: \(v^2 = u^2 + 2gh\).
Solution: Given \(u = 20\text{ m/s}\,\text{ }v = 80\text{ m/s}\,\text{ }g = 10\text{ m/s}^2\). Substituting these values: \(80^2 = 20^2 + 2(10)h\). \(6400 = 400 + 20h\). \(6000 = 20h\) => \(h = 300\text{ m}\).

Concept: Equations of motion under gravity for a specific interval.
Formula: \(h = ut + \frac{1}{2}gt^2\).
Solution: Let \(u\) be velocity at window top. Given \(h=1.5\text{ m}\,\text{ }t=0.1\text{ s}\,\text{ }g=10\text{ m/s}^2\). \(1.5 = u(0.1) + \frac{1}{2}(10)(0.1)^2\). \(1.5 = 0.1u + 0.05\). \(1.45 = 0.1u\) => \(u = 14.5\text{ m/s}\).

Concept: Distances covered by a freely falling body in equal successive time intervals.
Rule: For a body falling from rest, distances in successive equal time intervals are in ratio 1:3:5:...
Solution: \(h_1:h_2:h_3 = 1:3:5\). This implies \(h_2 = 3h_1\) and \(h_3 = 5h_1\). Therefore, \(h_1 = h_2/3 = h_3/5\).