Question

Q.13 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?​

Answer 1

Answer:

v=0

u=40m\s

g=-10

Explanation:

2as=v^2-u^2

s=0-1600/-20

s=80m

thus distance will be 2×80=160

but however the ball will come back to same point therefore it's displacement will be 0

Answer 2

$\huge\underline\mathbb {ANSWER:-}$

U = 40m/s

As the stone is thrown upward the acceleration due to gravity is to be taken negative.

G = -10 m/s²

V² - U² = 2as

For free fall we can write this equation as,

• V² - U² = 2gh

As the stone reaches the maximum height its final velocity V = 0.

Thus,

↪0 - (40)² = 2x (-10) x Height (h)

↪-1600 = -20 × Height (h)

↪Height = 80 m

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.