Question

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​

Answer 1

Answer:

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Explanation:

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