91Ó°ÊÓ

The maximum horizontal displacement covered by the projectile during its motion is called horizontal range. The velocity and angle of projection determine the horizontal range.

The expression for the horizontal range is,

$\mathit{R}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\left(2\mathrm{θ}\right)}{\mathbf{g}}$

Here $\mathit{R}$is the horizontal range, ${\mathit{v}}_{\mathbf{0}}$is the velocity of projection, $\mathit{θ}$is the angle of projection and$\mathit{g}$is the acceleration due to gravity.

• Speed of projection is,$v_0=50\text{m/s}$.
• The angle of projection is,${\mathrm{θ}}_1=15°$.
• The horizontal range is,$R_1=128 \text{m}$.
• The angle of projection is, ${\mathrm{θ}}_2=45°$.
• The horizontal range is, $R_2=255 \text{m}$.
• The angle of projection is, ${\mathrm{θ}}_3=75°$.
• The horizontal range is,$R_3=128 \text{m}$.

When the angle of projection is $15°$, the horizontal range can be calculated as,

$R_1=\frac{v_0^2\sin \left(2{\mathrm{θ}}_1\right)}{g}$

Substitute the value in the above expression, and we get,

$\begin{array}{rcl}{R}_1& =& \frac{{\left(50\text{m/s}\right)}^2×\sin \left(2×15°\right)}{\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ & ≈& 128\text{m}\\ & & \end{array}$

Hence, when the angle of projection is $15°$, the range of the projectile is $128\text{m}$.

When the angle of projection is $45°$, the horizontal range can be calculated as,

$R_2=\frac{v_0^2\sin \left(2{\mathrm{θ}}_2\right)}{g}$

Substitute the value in the above expression, and we get,

$\begin{array}{rcl}{R}_2& =& \frac{{\left(50\text{m/s}\right)}^2×\sin \left(2×45°\right)}{\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ & ≈& 255\text{m}\\ & & \end{array}$

Hence, when the angle of projection is $45°$, the range of the projectile is $255\text{m}$.

When the angle of projection is $75°$, the horizontal range can be calculated as,

$R_3=\frac{v_0^2\sin \left(2{\mathrm{θ}}_3\right)}{g}$

Substitute the value in the above expression, and we get,

$\begin{array}{rcl}{R}_3& =& \frac{{\left(50\text{m/s}\right)}^2×\sin \left(2×75°\right)}{\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ & ≈& 128\text{m}\\ & & \end{array}$

Hence, when the angle of projection is $75°$, the range of the projectile is $128\text{m}$.