Free Falling Motion

Question

A rock is dropped from a height of 256 ft and falls toward the earth in a straight line. In t seconds, the rock drops a distance of s = 16t^2 ft. Answer the following showing all work.

 

a.) When does the rock hit the ground?

b.) What is the average velocity of the rock when it is falling?

c.) Use one of the definitions of derivative provided below to find the instantaneous velocity of the rock when it hits the ground: Show all work.

v(t) = s'(t) = lim (w->t) (s(w)-s(t))/(w-t) or v(t) = s'(t) = lim (h->0) (s(t+h)-s(t))/(h)

Answer 1

s=-16t^2+256=0 

              -256   -256

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              -16t^2=-256 Divide both sides by -16

                   t^2=16 Take the square root of both sides

                       t=4 seconds

 

a) Hence it would take 4 seconds for the rock to hit the ground.

b)

      v_ave=(256-0)/(0-4)=256/-4

      v_ave=-64 ft/sec Hence, the average velocity of the rock is -64 ft/sec when falling.

 

c) s(t)=-16t^2+256

    s(t+h)=-16(t+h)^2+256

    s(t+h)=-16(t^2+2ht+h^2)+256

    s(t+h)=-16t^2-32ht-16h^2+256

 

v(t)=s'(t)=lim h→0 (s(t+h)-s(t))/h

v(t)=lim h→0 (-16t^2-32ht-16h^2+256-(-16t^2+256))/h

v(t)=lim h→0 (-16t^2-32ht-16h^2+256+16t^2-256)/h 

v(t)=lim h→0 (-32ht-16h^2)/h=lim h→0 h(-32t-16h)/h

v(t)=lim h→0 -32t-16h =-32t-16(0)

v(t)=-32t

v(4)=-32(4)=-128 ft/sec Hence, the instantaneous velocity of the rock upon hitting the ground is -128 ft/sec.