A projectile is fired from the ground with velocity 𝑣⃗ = (12𝑖⃗ + 24𝑗⃗ ) 𝑚/𝑠. a) What is the velocity after 4 𝑠? b) What is the position of the point at which the height is maximum? c) What is the horizontal distance?

a) To find the velocity after 4 seconds, we need to consider the projectile's initial velocity and the effects of gravity. The initial velocity 𝑣⃗ = (12𝑖⃗ + 24𝑗⃗ ) 𝑚/𝑠 can be split into horizontal and vertical components: The horizontal component is 𝑣⃗_𝑥 = 12𝑖⃗ 𝑚/𝑠. The vertical component is 𝑣⃗_𝑦 = 24𝑗⃗ 𝑚/𝑠. Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion. So, after 4 seconds, the horizontal velocity will still be 12𝑖⃗ 𝑚/𝑠. The vertical velocity changes due to the acceleration of gravity. The acceleration due to gravity is 𝑔 = -9.8𝑗⃗ 𝑚/𝑠^2. Using the equation of motion for vertical motion: 𝑣⃗_𝑦 = 𝑣⃗_𝑖𝑦 + 𝑔𝑡 Substituting the given values: 𝑣⃗_𝑦 = 24𝑗⃗ - 9.8𝑗⃗ (4) 𝑣⃗_𝑦 = 24𝑗⃗ - 39.2𝑗⃗ 𝑣⃗_𝑦 = (24 - 39.2)𝑗⃗ 𝑣⃗_𝑦 = -15.2𝑗⃗ 𝑚/𝑠 Therefore, the velocity after 4 seconds is 𝑣⃗ = (12𝑖⃗ - 15.2𝑗⃗ ) 𝑚/𝑠. b) To find the position of the point at which the height is maximum, we need to consider the vertical motion of the projectile. The height of the projectile can be calculated using the equation of motion for vertical motion: ℎ = ℎ_𝑖 + 𝑣⃗_𝑖𝑦𝑡 + 0.5𝑔𝑡^2 Since the projectile is fired from the ground, the initial height ℎ_𝑖 is zero. To find the time at which the height is maximum, we can use the fact that the vertical velocity becomes zero at the highest point. Setting 𝑣⃗_𝑦 = 0: 0 = 24 - 9.8𝑡 9.8𝑡 = 24 𝑡 = 24/9.8 𝑡 ≈ 2.45 𝑠 Substituting this time into the equation of motion: ℎ = 0 + (24)(2.45) + 0.5(-9.8)(2.45)^2 ℎ ≈ 59.22 𝑚 Therefore, the position of the point at which the height is maximum is approximately 59.22 meters above the ground. c) To find the horizontal distance, we need to consider the horizontal motion of the projectile. The horizontal distance can be calculated using the equation of motion for horizontal motion: 𝑥 = 𝑥_𝑖 + 𝑣⃗_𝑥𝑡 Since the projectile is fired from the ground, the initial horizontal position 𝑥_𝑖 is zero. Substituting the given values into the equation of motion: 𝑥 = 0 + (12)(4) 𝑥 = 48 𝑚 Therefore, the horizontal distance traveled by the projectile after 4 seconds is 48 meters.