a) i\'m standing by a 2nd floor window in a building. Someone on the ground belo

Question

a) i'm standing by a 2nd floor window in a building. Someone on the ground below throws a rock upward. I start a stopwatch when the rock passes going upward and stop the stopwatch when it passes coming back down. The time interval is 3s. What is the speed of the rock as it passes the window?

b) A ball is dropped (from rest) from the top of a building. I'm standing in front of a window that is 1.5m high. If it takes 0.2s for the ball to pass by the window, from what height above the top of the window was the ball dropped?

Explanation / Answer

a) let initial velocity = u Final velocity, v = 0 acceleration, a = -g = -9.8 m/s^2 Time interval, t = 3s we have an equation, v = u + at by substituting all the values we get, u = 29.4 m/s b) The initial velocity of the ball, u = 0 let the total distance = x m The height of the window = 1.5 m So the distance traveled by the ball, s = ( x - 1.5) m Time taken by the ball to travel this distance, t = 0.2 s Acceleration, a = 9.8 m/s^2 We have an equation, s = ut + 0.5at^2 from this, x = 0.5 * 9.8 * (0.2)^2 = 0.196 m So the height at which the ball is dropped above the window is = 0.196 m

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