Answer to Question #176350 in Physics for Aveline

Question

In preparation for the 2018 Asian Games, the Philippine shooting team is practicing almost every day. From the prone position, one of the shooters tries to aim at a target that is 50 m away. He places the rifle placed on his shoulder at about 1.5 m above the ground. How long it will take for the bullet to reach the target? With what speed should the bullet leave the muzzle of the riffle to reach the target?

Answer 1

Assuming that the target is located at the level of 0m, and the muzzle is at $h = 1.5m$ above it, the time $t$ it takes for the bullet to reach the ground in vertical movement is given by the kinematic equation:

$$h = \dfrac{gt^2}{2}\\ t = \sqrt{\dfrac{2h}{g}}\\ t = \sqrt{\dfrac{2\cdot 1.5}{9.8}}\approx 0.6s$$

where $g = 9.8m/s$ is the gravitational acceleration.

In this time the bullet should cover $d = 50m$ in the horizontal movement. In order to do this it should have the following speed:

$$v_0 = \dfrac{d}{t} = \dfrac{d\sqrt{2h}}{\sqrt{g}}\\ v_0\approx 90m/s$$

Answer. 0.6s, 90m/s.