Asked by Tom
If you launch a ball precisely 8' vertically how long does it take to go from the ground to 8' and back to the ground? If you then launch the same ball precisely 10' vertically, how much longer will it take to return to the ground?
Answers
Answered by
tchrwill
The initial velocity needed to launch the ball to a height of 8 feet derives from Vf^2 = Vo^2 - 2gh where Vo = the initial speed, Vf = the final speed, g = the acceleration due to gravity and h = the height reached .
Thus, 0 = Vo^2 - 2(32.2)8 making Vo = 22.7fps.
The time to reach the height of 8 feet derives from
Vf = Vo - gt or 0 = 22.7 - 32.2t making t = .7 seconds.
The time to fall back to the ground is exactly the same deriving from
h = Vot + gt^2/2 or 8 = 0(t) + 16.1t^2 yielding t = .7 seconds.
I'll let you doo the math for the 10 ft. scenario.
Thus, 0 = Vo^2 - 2(32.2)8 making Vo = 22.7fps.
The time to reach the height of 8 feet derives from
Vf = Vo - gt or 0 = 22.7 - 32.2t making t = .7 seconds.
The time to fall back to the ground is exactly the same deriving from
h = Vot + gt^2/2 or 8 = 0(t) + 16.1t^2 yielding t = .7 seconds.
I'll let you doo the math for the 10 ft. scenario.
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