A stone is thrown vertically upward with velocity of 300m/sec If acceleration due to gravity is 10m/sec what is distance travelled by particle during first second of motion
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Explanation:
ANSWER BY ADI5166
Initial Velocity u=40
Fianl velocity v=0
Height, s=?
By third equation of motion
v
2
−u
2
=2gs
0−40
2
=−2×10×s
s=
20
160
⇒s=80m/s
Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m
Total Diaplacement =0, Since the initial and final point is same.
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