Given, 
Length of the string, l = 80 cm
Number of revolutions = 14
Time taken = 25 s


The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

The parallelogram law of vector addition states that if two vectorsare represented by the two adjacent sides of a parallelogram by both magnitude and direction from a point, then their resultant can be represented by the diagonal of the parallelogramfrom the same point.

Let us consider a ball projected at an angle θ with respect to horizontal x-axis with the initial velocity u .

The point O is called the point of projection, θ is the angle of projection and OB = horizontal range. The total time taken by the particle from reaching O to B is called the time of flight.

Now,

(a). The total time of flight is

Resultant displacement is zero in Vertical direction.

Therefore, by using equation of motion

$$\begin{gathered} \mathrm{s} = \mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{gt}=2 \sin \mathrm{\theta } \\ \mathrm{t} = \frac{2 \sin \mathrm{\theta }}{\mathrm{g}} \end{gathered}$$

(b). The  horizontal range is

Horizontal range OA = horizontal component of velocity × total flight time

$$\begin{gathered} \mathrm{R} =\mathrm{u} \cos \mathrm{\theta } \times \frac{2\mathrm{u} \sin \mathrm{\theta }}{\mathrm{g}} \\ \mathrm{R} = \frac{{\mathrm{u}}^2 \sin 2\mathrm{\theta }}{\mathrm{g}} \end{gathered}$$

(c). The maximum height is

It is the highest point of the trajectory point A. When the ball is at point A, the vertical component of the velocity will be zero.

By using equation of motion

$$\begin{gathered} {\mathrm{v}}^2={\mathrm{u}}^2 -2\mathrm{as} \\ 0 = {\mathrm{u}}^2{\sin }^2\mathrm{\theta }-2\mathrm{gH} \\ \mathrm{H}= \frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}} \end{gathered}$$

Hence, this is the required solution

A physical quantity that requires only magnitude for its complete specification is called a scalar quantity.