Question

युयम् प्राग्या छात्रा: स्थ । युयम् स्वातिरिक्तसमये पुस्तकालयम् आगच्छत ​

Answer 1

Answer:

According to the equation of the motion under gravity

v

2

−u

2

=2gs

u=initial velocity of the stone=40m/s

v= Final velocity of the stone=0m/s

Let h be the maximum height attained by the stone

Therefore,

0

2

−40

2

=2(−10)h

h=(40×40)/20=80

Therefore, total distance covered by the stone during its upward and downward journey=80+80=160m

Net displacement during its upward and downward journey=80+(−80)=0