Question

A stone is thrown vertically upwards with
initial velocity of 40 m/s. Taking g = 10 m/s²
find the maximum and total distance covered
by stone.
(s = 160 m, distance = 0)​

Answer 1

Answer:

As the stone reaches the maximum height its final velocity V = 0. So, the maximum height to which the stone can reach is 80 m. Total distance covered by the stone = 80 + 80 = 160 m. And, as the stone comes back to its initial position the displacement of the stone = 0.

Explanation:

According to the equation of the motion under gravity v² −u² =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,

0² −40² =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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