Let's say there is this situation: A load of bricks is being lifted by a crane at the steady velocity of 5 m/s when one brick falls off 6m above the groundWhat is the greatest height the brick reaches above ground, how long does it take to reach the ground, what is the speed just before it hits the ground? Now that was the question, but I would like to know the concept behind this: It seems a bit too easy for the greatest height the brick has reached to be 6 m.but the brick falls off at 6mDoes this mean that it goes a bit higher than 6m and then falls, or just simply falls at 6 m? Please help!
Well you see, the brick can't simply 'fall' once it loses contact with the other bricks because it is already moving UPWARDSWhat happens is that the brick keeps moving upwards with the initial speed of 5m/sIt can't just suddenly fall because the brick's inertia (resistance to changes in motion) carries it forwardThe brick reaches a maximum height above the position where it was when it initially lost contact, and then dropsSo to find the maximum height it can reach, simply apply the equations of motions as you've probably done many times before: The one you need is v^2 - u^2 2as u 5 m/s v 0 m/s (at the maximum, the brick will stop moving) a -9.8 m/s^2 (DON'T forget the minus sign) t don't care s what we want to know So just rearrange for sThis then tells us the extra height the brick travels after it loses contactI calculated it to be 1.28m So now, to get the maximum height above ground, you simply add 6 m (the height above ground) to thisSo the final answer is 7.28mI hope this helps!
Well you see, the brick can't simply 'fall' once it loses contact with the other bricks because it is already moving UPWARDSWhat happens is that the brick keeps moving upwards with the initial speed of 5m/sIt can't just suddenly fall because the brick's inertia (resistance to changes in motion) carries it forwardThe brick reaches a maximum height above the position where it was when it initially lost contact, and then dropsSo to find the maximum height it can reach, simply apply the equations of motions as you've probably done many times before: The one you need is v^2 - u^2 2as u 5 m/s v 0 m/s (at the maximum, the brick will stop moving) a -9.8 m/s^2 (DON'T forget the minus sign) t don't care s what we want to know So just rearrange for sThis then tells us the extra height the brick travels after it loses contactI calculated it to be 1.28m So now, to get the maximum height above ground, you simply add 6 m (the height above ground) to thisSo the final answer is 7.28mI hope this helps!
