i) Vf = Vi + g * t
0 = 32 - 10 * t
t = 0.32 seconds ( 320 milliSeconds)
ii) DeltaY= H = Vi * t + 0.5 * 9.8 * 0.32 ^ 2
H = 9.73 m
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Question 4: Is not too clear ,However , if Maxima displacement is required apply eqn II :
Vi = 1000 m/s
Vf =0
H =?
g = -10 m/s^2
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Question 5:
Finding displacement from point of release :
Apply eqn I)
H1= -10 +0.5(-10)(10)^2
= 510 m
Note Vi = -1m/s (1 m/s downwards) reason is for the defination of displacement
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