Question

A modern riffle can fire a bullet from 180 m/s to 1500 m/s.

a. Compute for the maximum height of a bullet fired by a riffle with an average velocity of 500m/s at an angle of 800.

Answer 1

Let's write the equation of motion of the bullet in vertical direction:

$$y=v_0tsin\theta-\dfrac{1}{2}gt^2, (1)$$

here, $v_0 = 500\ \dfrac{m}{s}$ is the initial velocity of the bullet, $t$ is the time of flight of the bullet, $\theta=80^{\circ}$ is the launch angle, $y$ is the vertical displacement of the bullet (or the height) and $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's first find the time that the bullet takes to reach the maximum height from the kinematic equation:

$$v_y=v_0sin\theta-gt_{rise},$$ $$0=v_0sin\theta-gt_{rise},$$ $$t_{rise}=\dfrac{v_0sin\theta}{g}.$$

Then, we can substitute $t_{rise}$ into the first equation and find the maximum height:

$$y_{max}=v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(500\ \dfrac{m}{s})^2\cdot sin^280^{\circ}}{2\cdot 9.8\ \dfrac{m}{s^2}}=12370\ m=12.37\ km.$$

Answer:

$y_{max}=12370\ m=12.37\ km.$