Physics, asked by nilugupta7777, 7 months ago

A stone is thrown vertically upwards from the
ground and caught at the same point on the
ground after 10 seconds. (Take, g = 10 m/s2)
The maximum height reached by stone is​

Answers

Answered by ankitathakur1604
0

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

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