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?

Here's how to solve this problem:<br /><br />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):<br /><br /> KE = (1/2) * m * v^2<br /><br /> Solving for m:<br /><br /> m = (2 * KE) / v^2<br /> m = (2 * 284 J) / (240 m/s)^2<br /> m = 568 J / 57600 m²/s²<br /> m ≈ 0.00986 kg (approximately 9.86 grams)<br /><br />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):<br /><br /> v² = v₀² + 2 * g * Δy<br /><br /> 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:<br /><br /> 0 = (240 m/s)² + 2 * (-9.8 m/s²) * Δy<br /><br /> Solving for Δy:<br /><br /> Δy = (240 m/s)² / (2 * 9.8 m/s²)<br /> Δy = 57600 m²/s² / 19.6 m/s²<br /> Δy ≈ 2939 m<br /><br />Therefore, the bullet will travel approximately 2939 meters high.<br />