18.A stone is thrown vertically upwards with an initial velocity of 40m/s. Find the max height reached by the stone. What is the net displacement and the total distance covered by the stone?
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Explanation:
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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