Question

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

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Answer 1

Answer:

your answer in the attachment

Answer 2

Explanation:

Initial Velocity u= 40

Fianl velocity v= 0

Height, s= ?

By third equation of motion

${v}^{2} - {u}^{2} = 2gs \\ 0 - {40}^{2} = - 2 \times 10 \times s \\ s = \frac{160}{20}$

⇒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.