Question #169610

An object falls from rest, and s =-16t² where s feet is the distance of the object from the starting point at t seconds, and the positive direction is upward. If a stone is dropped from a building 256 ft high, find


A. the instantaneous velocity of the tone 1 sec after it is dropped;


B. the instantaneous velocity of the stone 2 sec after it is dropped;


C. how long it takes the stone to reach the ground;


D. the instantaneous velocity ot the stone when it reaches the ground.


1
Expert's answer
2021-03-09T07:02:28-0500

Given,

s=16t2s= -16t^2\\

g=32ft/sec2g= 32ft/sec^2

S=256ftS= 256ft


(a) Instantaneous velocity of the stone= g×tg\times t ft/sec

So, at t=1sec, the instantaneous velocity of stone = g×t=32×1=32ft/secg\times t=32\times 1=32ft/sec


(b) Similarly, at t=2sec, the instantaneous velocity of stone =g×t2sec=32×2=64ft/secg\times t_{2sec}=32\times 2= 64 ft/sec


(c) Total distance travelled to reach the ground S=12gt2S=\dfrac{1}{2}gt^2

So, total time it takes to reach ground t=2Sg=2×25632=16=4sect=\sqrt{\dfrac{2S}{g}}=\sqrt{\dfrac{2\times256}{32}}=\sqrt{16}=4 sec


(d) The stone reaches the ground at t=4sec

So, instantaneous velocity of stone when it reaches ground

v=g×t4sec=32×4=128ft/secv= g\times t_{4sec}=32\times 4=128ft/sec


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