A stone is thrown vertically upward with velocity 40m/s. Find the maximum height reached by the stone .What is the the net displacement and total distance covered by the stone.
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Answer: distance-160m, displacement- 0m
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= 160 divided by 20
⇒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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