person is watching through a window of an apartment sees a ball that rises vertically upward and then vertically down for a total time of 0.5 seconds. if the height of the window is 2 metre, find the max height above the window reached by the ball
To find the maximum height above the window reached by the ball, we need to first understand the motion of the ball.
The ball rises vertically upward and then vertically downward, completing one full cycle. The total time for this cycle is given as 0.5 seconds.
Since the ball goes up and then comes back down, we can assume that the time taken to reach the maximum height is half of the total time for the entire cycle. So, the time taken to reach the maximum height is 0.5 seconds divided by 2, which is 0.25 seconds.
We can use this time to find the velocity at maximum height using the equation of motion: v = u + at, where
v = final velocity, u = initial velocity, a = acceleration, and t = time.
Since we are dealing with vertical motion, the acceleration due to gravity (g) needs to be considered. The ball is moving upward against gravity, so the acceleration should be negative.
At the highest point of the ball's motion, the velocity is zero. So, the equation becomes: 0 = u + (-g)(0.25).
Simplifying the equation, we find: u = 0.25g.
Now, we need to find the displacement (change in height) at the maximum height. Again, using the equation of motion: s = ut + 0.5at^2, where s = displacement.
At the highest point, the final velocity is zero, so the equation becomes: s = (0.25g)(0.25) + 0.5(-g)(0.25)^2.
Simplifying the equation, we find: s = 0.03125g.
Finally, to find the maximum height above the window reached by the ball, we subtract the height of the window (2 meters) from the displacement:
Maximum height = 0.03125g - 2 meters.
The value of 'g' is approximately 9.8 m/s^2 (acceleration due to gravity), so substituting this value, we can calculate the maximum height reached by the ball above the window.