A rifle is aimed horizontally at a target 30maway. The bullet hits the target 1.9 cm below the aiming point. What are (a) the bullet’s time of flight and (b) its speed as it emerges from the rifle?

1. Horizontal distance, x=30m
2. Vertical distance at which the bullet hits the target,${\mathrm{y}}_0=1.9\mathrm{cm}\mathrm{or}0.019\mathrm{m}$

Using the equation for the position, we can find the velocity in terms oft by differentiating it. Using the equation for the acceleration of the second particle, we can find the equation for the velocity for the second particle, by integrating this. Finally, we can equate the two equations to find the time.

Formulae:

The second horizontal equation of kinematic motion, $\mathrm{y}-{\mathrm{y}}_0={\mathrm{V}}_{0\mathrm{y}}\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2$ …(i)

The velocity of a body in motion, ${\mathrm{V}}_{\mathrm{x}}=\frac{\mathrm{x}-{\mathrm{x}}_0}{\mathrm{t}}$ …(ii)

Rearranging the equation (i) for the time and assuming initial vertical velocity as zero, we get the value of time of flight of the bullet as follows:

$$\begin{gathered} t=\sqrt{\frac{-2\left(y-y_0\right)}{g}} \\ =\sqrt{\frac{-2\left(0m-0.019m\right)}{9.8m/s^2}} \\ =6.2\times {10}^{-2}s \end{gathered}$$

Hence, the value of time of the bullet is$6.2\times {10}^{-2}s$.

For the initial velocity, we have no acceleration, so we can use the equation of constant velocity. Thus, using the given in equation (ii), we get the initial speed of the bullet as:

$$\begin{gathered} V_x=\frac{30m-0m}{6.2\times {10}^{-2}s} \\ =483.87m/s \\ =4.8\times {10}^{2}m/s \end{gathered}$$

Hence, the value of the velocity is$4.8\times {10}^2\mathrm{m}/\mathrm{s}$.