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where
Δ
v
is the change in its velocity during contact with the ground and
3
20.0 10 s
t
−
Δ=
×
is the duration of contact. Now, to find the velocity just
before
contact,
we put the origin at the point where the ball is dropped (and take +
y
upward) and take
t
=
0 to be when it is dropped. The ball strikes the ground at
y
= –15.0 m. Its velocity there is
found from Eq. 216:
v
2
= –2
gy
. Therefore,
2
2
2(9.8 m/s )( 15.0 m)
17.1 m/s
vg
y
=− −
−
=−
where the negative sign is chosen since the ball is traveling downward at the moment of
contact. Consequently, the average acceleration during contact with the ground is
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics, Acceleration

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