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?

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?

Answer 1

Here's how to solve this problem:1. **Find the mass of the bullet:** We know the kinetic energy (KE) and the initial velocity (v) of the bullet. We can use the kinetic energy formula to find the mass (m): KE = (1/2) * m * v^2 Solving for m: m = (2 * KE) / v^2 m = (2 * 284 J) / (240 m/s)^2 m = 568 J / 57600 m²/s² m ≈ 0.00986 kg (approximately 9.86 grams)2. **Find the maximum height:** At the maximum height, the bullet's velocity will be zero. We can use the following kinematic equation to relate initial velocity (v₀), final velocity (v), acceleration due to gravity (g), and displacement (Δy, which represents the height in this case): v² = v₀² + 2 * g * Δy Since the final velocity (v) is zero at the highest point, and the acceleration due to gravity (g) is -9.8 m/s² (negative because it acts downwards), we have: 0 = (240 m/s)² + 2 * (-9.8 m/s²) * Δy Solving for Δy: Δy = (240 m/s)² / (2 * 9.8 m/s²) Δy = 57600 m²/s² / 19.6 m/s² Δy ≈ 2939 mTherefore, the bullet will travel approximately 2939 meters high.

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

Answer