Question

A 1 kg projectile is fired from a cannon with an initial kinetic energy of 104 J. The cannon has an elevation angle of 45◦. How far does the projectile go before striking the ground (neglect the air resistance)?

Answer 1

The projectile travels approximately 10.6 meters before striking the ground.

Step-by-step explanation:

To determine the distance the projectile travels, we need to use the equation for horizontal distance: d = v0x * t, where v0x is the initial horizontal velocity and t is the time of flight. We can find v0x using the equation v0x = v0 * cos(θ), where v0 is the initial velocity and θ is the angle of elevation. Since the initial kinetic energy is given, we can use the formula KE = 0.5 * m * v02 to find v0. Solving for t in the vertical motion equation y = v0y * t - 0.5 * g * t2 gives us the time of flight.

Using the given values, we have:

v0 = √(2 * KE / m) = √(2 * 104 / 1) = √208 ≈ 14.4 m/s

v0x = v0 * cos(θ) = 14.4 * cos(45◦) ≈ 10.2 m/s

v0y = v0 * sin(θ) = 14.4 * sin(45◦) ≈ 10.2 m/s

t = v0y / g = 10.2 / 9.8 ≈ 1.04 s

Finally, substituting the values into the horizontal distance equation, we have:

d = v0x * t = 10.2 * 1.04 ≈ 10.6 m