A projectile is thrown at an angle θ from the horizontal with velocity u under the gravitation field of the earth. Derive expressions for its:
a)Time of its flight
b)Height
c)Horizontal Range

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