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A rock is tossed straight up with a velocity of 25m/s When it returns, it falls into a hole 10m deep.

a)What is the rock's velocity as it hits the bottom of the hole?

b)How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

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Added Wed, 09 Sep '15
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final velocity v =0
initial speed = 25 m/s

time taken to reach max. height = t1

v = u-gt1
0 = 25-9.8t1
t1 = 25/9.8 = 2.55 seconds

Now we have,
max. height reached = s1
v^2 =y^2 -2gs
0 = 25^2 - 2 x9.8 x s
s = 31.888 m

Now total distance on way back s'= 31.888 + 10 = 41.888 m

now we have,
v^2 = u^2 +2gs'
= 0 + 2 x 9.8 x 41.888
Final Velocity v = 28.653 m/s

Also, v= u+gt2
28.653 = 0+ 9.8t2
t2 = 2.924 sec

Total time taken t = t1+ t2 = 2.55 + 2.924 = 5.474 seconds
Edit
Added Wed, 09 Sep '15
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