You drop a 2.70 kg book to a friend who stands on the ground at distance D = 14.

Question

You drop a 2.70 kg book to a friend who stands on the ground at distance D = 14.0 m below. Your friend's outstretched hands are at distance d = 1.70 m above the ground (Fig. 8-29). (a) What is the speed of the book when it reaches her hands? (b) If we substituted a second book with twice the mass, what would its speed be when it reaches her hands? (c) If, instead, the book were thrown down with initial speed 6.40 m/s, what would be the answer to (a)?
http://edugen.wileyplus.com/edugen/courses/crs1650/art/qb/qu/c08/fig08_29.gif

Explanation / Answer

a)s = 0.5*a*t^2
a=g=9.8m/s^2
(14-1.7 ) = 0.5*9.8*t^2
t = 1.584s
speed =a*t = 15.53 m/s
b) even if we double the mass the speed is the same as accelaration does not depend on mass

speed =15.53m/s
c)
u= 6.4
s =u*t + 0.5*a*t^2
(14-1.7 ) = 6.4*t + 0.5*9.8*t^2
t = 1.06s
speed = 6.4 + 9.8*t = 16.8 m/s

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