Question

Verify the ranges shown for the projectiles in Figure 3.41 (b) for an initial velocity of $\mathbf{50}\mathbf{}\mathbf{\text{m/s}}$at the given initial angles.

Short answer

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

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

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

Step-by-step solution

Horizontal range

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,

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

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

Given data

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

Case (i) Horizontal range

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

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

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

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

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

Case (ii) Horizontal range

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

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

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

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

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

Case (iii) Horizontal range

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

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

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

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

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