An object is dropped from rest at a height of 115 m. Find the distance it falls

Question

An object is dropped from rest at a height of 115 m. Find the distance it falls during its finalsecond in the air.

I am getting confused by what they mean by final second. I triedusing the equation x=ViT + .5AT2 usingx as 115 and Vi as 0 and then solving for T. Once I got that isubtracted one from it and then used the same equation to solveback for x. The answer was wrong, so I don't know what i doingwrong

Explanation / Answer

If you use: x = (Vi)(t) + (.5)(a)(t^2) -115 = (0)(t) + (.5)(-9.8)(t^2) t should come out to be 4.8445 seconds. This means that the ball dropped for 4.8445 seconds. During it'sfinal second, 3.8445 seconds have passed already. Now you use Vf = Vi + (a)(t) ==> Vf = 0 + (-9.8)(3.8445) ===>Vf = 37.676 meters per second after 3.8445 seconds of freefall. Now your new Vi to finish the problem is 37.676. Use x = (Vi)(t) +(.5)(a)(t^2) again. This time, it's: x = (-37.676)(1) + (.5)(-9.8)(1^2) ===> x = -42.576 meters. It dropped 42.576 meters in the final second. The reason you got it wrong the first time is because when yousubtracted 1 from the time and put it back into the equation, youneglected the fact that Vi is no longer 0. Vi was the velocity onesecond before it hit the ground, which was not 0 anymore. Please rate lifesaver if I did this correctly. Thanks!

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