91Ó°ÊÓ

Curve of Range vs initial angle of projection

It deals with the kinematic equations of motion in which the motion can be described with constant acceleration. Also, this problem is based on the concept of projectile motion. When a particle is thrown near the earth’s surface, it travels along a curved path under constant acceleration directed towards the center of the earth surface. This is a projectile motion problem, which contains two independent motions.

Horizontal motion with constant speed ${\mathbf{a}}_{\mathbf{x}}\mathbf{=}\mathbf{0}$

Vertical motion with acceleration due to gravity $\mathbf{a}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$.

Formulae:

$$\begin{gathered} \mathrm{Range}\left(R\right)=v_{\mathit{0}x}T \\ {v}_{0x}=\mathrm{Horizontal}\mathrm{component}\mathrm{of}\mathrm{initial}\mathrm{speed} \\ T-\mathrm{Time}\mathrm{of}\mathrm{flight} \end{gathered}$$

We know that,

$\begin{array}{l}R=V_0{x}^T\\ =V_0×\cos \left({\mathrm{θ}}_0\right)×T\end{array}$

Hence,

$T=\frac{R}{V_0\cos \left({\mathrm{θ}}_0\right)}$

As$\mathrm{θ}$increases, time of flight also increases, value of$cos\left(\mathrm{θ}\right)$would decrease. Therefore, rank according to time is:

$\mathrm{Rank}:c\mathit{>}b\mathit{>}a$

At maximum height vertical component of velocity is zero, so speed is same as horizontal component of velocity which is same at any point.

Hence,

$\begin{array}{l}v=V_{0}x\\ =V_0×\cos \left({\mathrm{θ}}_0\right)\end{array}$

As$\mathrm{θ}$increases, speed decreases. Therefore, rank according to speed at maximum height is:

$\mathrm{Rank}:a>b>c$