A brick is dropped rest in the cliff. What is the speed of the brick in 5 seconds later?What is the velocity of the ball at this time? How far does it travel during the time? What the placement of the ball?draw an arbitrary graph!body vs timevelocity vs timeacceleration vs time
Answer (1)
Answer:The problem involves a brick dropped from rest in a cliff. We need to find its speed, velocity, distance traveled, and placement after 5 seconds.Assumptions: * The brick is dropped from rest, so its initial velocity (v₀) is 0 m/s. * The only force acting on the brick is gravity, which gives it a constant acceleration (a) of 9.81 m/s² (assuming no air resistance).Calculations:1. Speed (v) after 5 seconds:Using the equation of motion: v = v₀ + atv = 0 + 9.81 * 5v = 49.05 m/s2. Velocity (v) after 5 seconds:Velocity is a vector quantity that includes both magnitude (speed) and direction. Since the brick is falling downwards, its velocity is -49.05 m/s (negative sign indicates downward direction).3. Distance traveled (s) in 5 seconds:Using the equation of motion: s = v₀t + 0.5at²s = 0 * 5 + 0.5 * 9.81 * 5²s = 122.625 m4. Placement of the brick:Assuming the cliff's base is at s = 0, the brick's placement after 5 seconds is 122.625 m below the cliff's edge.Graphs:Body vs Time:Velocity vs Time:Acceleration vs Time:Note: The negative values in the velocity and acceleration graphs indicate downward direction.Therefore, after 5 seconds, the brick has a speed of 49.05 m/s, a velocity of -49.05 m/s, has traveled 122.625 m, and is located 122.625 m below the cliff's edge.sorry I cannot make arbitrary graph!so this maybe a helpful?let me know quick!
