Question

a rifle is fired horizontally and travels 200.0 m e. the rifle barrel is 1.90 m from the ground. what speed must the bullet have been travelling at? ignore friction.

Explanation:

Step1: Analyze vertical - motion

The bullet is in free - fall in the vertical direction. The initial vertical velocity $v_{0y}=0$ m/s, the acceleration due to gravity $g = 9.8$ m/s², and the vertical displacement $y=- 1.90$ m (taking downwards as negative). Use the equation $y=v_{0y}t+\frac{1}{2}at^{2}$. Since $v_{0y} = 0$ m/s, the equation simplifies to $y=\frac{1}{2}at^{2}$.
$y=\frac{1}{2}gt^{2}$

Step2: Solve for time $t$

Rearrange the equation $y=\frac{1}{2}gt^{2}$ to solve for $t$.
$t=\sqrt{\frac{-2y}{g}}$
Substitute $y = - 1.90$ m and $g = 9.8$ m/s² into the formula:
$t=\sqrt{\frac{-2\times(-1.90)}{9.8}}\approx\sqrt{\frac{3.8}{9.8}}\approx0.62$ s

Step3: Analyze horizontal - motion

In the horizontal direction, there is no acceleration ($a_x = 0$ m/s²), and the horizontal displacement $x = 200.0$ m. The horizontal velocity $v_x$ is constant, and we use the equation $x = v_x t$.
$v_x=\frac{x}{t}$

Step4: Calculate the horizontal velocity

Substitute $x = 200.0$ m and $t\approx0.62$ s into the formula:
$v_x=\frac{200.0}{0.62}\approx322.6$ m/s

Answer:

322.6 m/s