Question
7. Schwab fires a handgun straight upwards into the air (very foolishly). If the bullet leaves the muzzle of the gun with a velocity of 240m/s and a kinetic energy of 284 J, how high will the bullet travel?
Solution
4
(345 Votes)
Eliza
Professional · Tutor for 6 years
Answer
Here's how to solve this problem:**1. Understand the Concepts*** **Conservation of Energy:** As the bullet travels upwards, its kinetic energy is converted into potential energy. At the highest point, all kinetic energy becomes potential energy.* **Kinetic Energy:** KE = (1/2)mv²* **Potential Energy:** PE = mgh* **Gravity:** The acceleration due to gravity (g) is approximately 9.8 m/s² downwards.**2. Find the Mass of the Bullet**We are given the initial kinetic energy and velocity. We can use the kinetic energy formula to find the mass (m) of the bullet:KE = (1/2)mv²28.1 J = (1/2)m * (240 m/s)²56.2 J = m * 57600 m²/s²m = 56.2 J / 57600 m²/s²m ≈ 0.000976 kg (approximately 1 gram, which is reasonable for a bullet)**3. Calculate the Maximum Height**At the highest point, all the initial kinetic energy is converted to potential energy:KE_initial = PE_final(1/2)mv² = mghNotice that the mass (m) cancels out on both sides:(1/2)v² = ghNow, solve for h (height):h = v² / (2g)h = (240 m/s)² / (2 * 9.8 m/s²)h = 57600 m²/s² / 19.6 m/s²h ≈ 2939 m**Answer:** The bullet will travel approximately 2939 meters high.**Important Note:** This calculation ignores air resistance. In reality, air resistance would significantly reduce the maximum height reached by the bullet.