To find the resultant velocity using unit vectors:

- Initial velocity: \( \vec{u} = 0.4 \, \hat{i} \, \text{m/s} \)
- Acceleration: \( \vec{a} = 0.15 \, \hat{j} \, \text{m/s}^2 \)
- Time: \( t = 2 \, \text{seconds} \)

Using the equation of motion \( \vec{v} = \vec{u} + \vec{a}t \):

- In the \( x \)-direction: \( v_x = 0.4 \, \text{m/s} \)
- In the \( y \)-direction: \( v_y = 0 + 0.15 \times 2 = 0.3 \, \text{m/s} \)

The resultant velocity is:

\[
|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{(0.4)^2 + (0.3)^2} = \sqrt{0.25} = 0.5 \, \text{m/s}
\]

Thus, the resultant velocity is 0.5 m/s

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.

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

To find the acceleration, we need to compute the second derivatives of

$x$

and

$y$

with respect to

$t$

.

1. 
   $x = 7t + 4t^2$ 

   
   

   • First derivative (velocity in x-direction):
     $$\frac{dx}{dt}=7+8t$$ 
   • Second derivative (acceleration in x-direction):
     $$\frac{d^2x}{dt^2} = 8 \, \text{m/s}^2$$ 

     
     

2. 
   $y = 5t$ 
   • First derivative (velocity in y-direction):
     $$\frac{dy}{dt}=5$$ 
   • Second derivative (acceleration in y-direction):
     $$\frac{d^{2}y}{d{t}^2}=0{\text{m/s}}^2$$ 

Now, the total acceleration is given by:

$$a=\sqrt{(a_x{)}^2+(a_y{)}^2}=\sqrt{(8{)}^2+(0{)}^2}=8{\text{m/s}}^2$$

Thus, the acceleration of the particle at

$t = 5 \, \text{s}$

is

$8 \, \text{m/s}^2$

We Know that,

\[ \frac{R}{H}=\frac{4}{tan\theta}\]

\[ tan \theta = 4/3 \]

\[ H = \frac{u^{2}sin^{2}\theta}{2g} \]

Solving we get

\[ u= 5\sqrt{g/2}\]

The equation of motion of the projectile is given as:

\[
y = 12x - \frac{3}{4}x^2
\]

At the range, \( y = 0 \). To find the range, set \( y = 0 \) and solve for \( x \):

\[
0 = 12x - \frac{3}{4}x^2
\]

Multiply the entire equation by 4 to eliminate the fraction:

\[
0 = 48x - 3x^2
\]

Factor the equation:

\[
0 = x(48 - 3x)
\]

This gives two solutions:

\[
x = 0 \quad \text{or} \quad 48 - 3x = 0
\]

Solving for \( x \):

\[
x = \frac{48}{3} = 16
\]

At the highest point of its trajectory, the vertical component of the velocity becomes zero, while the horizontal component remains unchanged.

Initial kinetic energy \( K \) is given by:

\[
K = \frac{1}{2} m u^2
\]

At the highest point, only the horizontal component of the velocity \( u \cos 60^\circ \) remains. Therefore, the kinetic energy at the highest point is due to this horizontal velocity:

\[
K_{\text{highest}} = \frac{1}{2} m (u \cos 60^\circ)^2
\]

Since \( \cos 60^\circ = \frac{1}{2} \):

\[
K_{\text{highest}} = \frac{1}{2} m \left(\frac{u}{2}\right)^2 = \frac{1}{4} \times \frac{1}{2} m u^2 = \frac{K}{2}
\]

Thus, the kinetic energy at the highest point is \( \frac{K}{2} \).