A ball is dropped from the top of a building. The ball falls 3 times farther during its last second of freefall than it did during its third second of freefall. The ball fell 4.9 meters during its first second, how tall is the building and how much time is the ball in the air?
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We can use equation of motion to do this problem. Since it has not given that the ball falls under gravity we have to check whether it really falls under gravity or other mean.
`darr s = ut+1/2at^2`
`4.9 = 0+1/2xxaxx1^2`
`a = 9.8`
This acceleration is almost same as acceleration due to gravity.
So now we can sure that the ball falls under gravity.
`darrv = u+at` for the 3rd second.
`v = 0+9.8xx3`
`v_1 = 29.4m/s`
So at the last second of free fall the velocity would be;
`v_2 = 3xx29.4 = 88.2m/s`
`darrv^2 = u^2+2aS` for the whole motion.
`88.2^2 = 0+2xx9.8xxS`
`S = 396.9m`
`darrv = u+at` for the whole motion.
`88.2 = 0+9.8xxt`
`t = 9s`
So the building is 396.9m tall and the ball is in air for 9 seconds.
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