Question

Gun sights are adjusted to aim high to compensate for the effect of

gravity, effectively making the gun accurate only for a specific range.

(a) If a gun is sighted to hit targets that are at the same height as the gun and$100.0$m away, how low will the bullet hit if aimed directly at a target $150.0$m away? The muzzle velocity of the bullet is$275$m/s.

(b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

Short answer

(a)The change in a vertical position during the $150.0$m flight, ∆y = $0.480$-m

(b)Higher the muzzle velocity, the smaller the deviation in the vertical direction. Air resistance slows the bullet down.

Step-by-step solution

Definition of the final velocity

When gravity first exerts a force on an item, its initial velocity indicates how fast it travels. The final velocity, on the other hand, is a vector number that measures a moving body's speed and direction after it has reached its maximum acceleration.

Given data:

• The horizontal distance between the gun and target, R=$100.0$m.
• The muzzle velocity, v_i = $275$m/s.

(a) To calculate the change in a vertical position during the 150 m flight

We have the equation for the range,

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

Substituting the values in the above the equation, we get

$\begin{array}{rcl}100\text{m}& =& \frac{{(275\text{m/s})}^2\sin \left(2{\theta }_i\right)}{9.8{\text{m/s}}^{\text{2}}}\\ 980\text{m}& =& 75600\text{m}\times \sin \left(2{\theta }_i\right)\\ \sin \left(2{\theta }_i\right)& =& \frac{980}{75600}\\ \sin \left(2{\theta }_i\right)& =& 0.0130\\ 2{\theta }_i& =& {\sin }^{-1}(0.0130)\\ 2{\theta }_i& =& 0.745\\ {\theta }_i& =& 0.373°\\ & & \end{array}$

Step 3: To calculate the horizontal component

Now we have to find the horizontal component of the velocity,

From the figure,

$\begin{array}{rcl}\cos \theta & =& \frac{v_{ix}}{v_i}\\ v_{ix}& =& v_i\cos \theta \\ & & \end{array}$ ………………..(1)

Substituting the values in equation 1, we get

$\begin{array}{rcl}{v}_{ix}& =& 275\times \cos (0.373)\\ & =& 274.99m/s\\ & & \end{array}$
To find the time required to reach the target, we have the formula for average velocity,

$\overline{v}=\frac{x}{t}$

Here, the average velocity = velocity along the x-direction

Therefore,

role="math" localid="1668667614123" $\begin{array}{rcl}t& =& \frac{x}{\overline{v}}\\ & =& \frac{x}{v_{ix}}\\ & =& \frac{100}{274.99}\\ & =& 0.364\text{s}\\ & & \end{array}$

If the target's horizontal distance is 150 m, the time is taken,

$\begin{array}{rcl}t& =& \frac{x}{v_{ix}}\\ & =& \frac{150}{274.99}\\ & =& 0.545\text{s}\\ & & \end{array}$

Here from the figure,

$\begin{array}{rcl}\sin \theta & =& \frac{v_{iy}}{v_i}\\ v_{iy}& =& v_i\sin \theta \\ & & \end{array}$

$\begin{array}{rcl}{v}_{iy}& =& \sin (0.373)\times 275\\ v_{iy}& =& 1.79\text{m/s}\\ & & \end{array}$

Finally, to calculate the change in a vertical position during the 150 m flight, we have the equation,

$\mathit{\Delta }\mathit{y}\mathbf{=}{\mathit{v}}_{\mathbf{i}\mathbf{y}}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{a}}_{\mathbf{y}}{\mathit{t}}^{\mathbf{2}}$

Therefore equation 2 becomes,

$\begin{array}{rcl}\Delta y& =& 1.79\times 0.545+\frac{1}{2}\times (-9.8)\times (0.545{)}^2\\ \Delta y& =& -0.480\text{m}\\ & & \end{array}$

The negative sign shows that the bullet will fall below the level of release.

(b) Discuss how a larger muzzle velocity and air resistance affect this problem.

Higher the muzzle velocity, the smaller the deviation in the vertical direction because the time of flight would be shorter. Air resistance would have the effect of decreasing the time of flight, which will obviously slow the bullet down.