Solution:

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 \( 18 \, \text{m} \).

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Solution:

The time taken by an object to fall from a height is given by the equation:

\[
t = \sqrt{\frac{2h}{g}}
\]

For mass \( m_a \) dropped from height \( a \), the time taken is:

\[
t_a = \sqrt{\frac{2a}{g}}
\]

For mass \( m_b \) dropped from height \( b \), the time taken is:

\[
t_b = \sqrt{\frac{2b}{g}}
\]

The ratio of time taken by the two bodies is:

\[
\frac{t_a}{t_b} = \frac{\sqrt{\frac{2a}{g}}}{\sqrt{\frac{2b}{g}}} = \sqrt{\frac{a}{b}}
\]

So, the ratio of the time taken is:

\[
\frac{t_a}{t_b} = \sqrt{\frac{a}{b}}
\]

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Solution:

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 \( \frac{5h}{3} \).

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Solution:

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.

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

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Solution:

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 

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Solution:

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

solving we get, t =2.5 sec.

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

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Solution:

Effective accelerations:
- Ascent: \( a_{\text{eff}} = g + a = 10 + 2 = 12 \, \text{m/s}^2 \)
- Descent: \( a'_{\text{eff}} = g - a = 10 - 2 = 8 \, \text{m/s}^2 \)

Ratio of times:
\[
\frac{t_a}{t_d} = \sqrt{\frac{a'_{\text{eff}}}{a_{\text{eff}}}} = \sqrt{\frac{8}{12}} = \sqrt{\frac{2}{3}}
\]

The ratio of time of ascent to time of descent is √(2/3).

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Solution:

\[ t =\frac{2u}{g}=\frac{2\times 30}{10}= 6 sec\]

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Solution:

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.

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Solution:

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

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

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

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