Question

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

Answer 1

Given :-

• Initial velocity (u) = 40m/s

• Final velocity (v) = 0m/s

• Acceleration due to gravity (g) = 10m/s²

To Find:-

• Maximum height of stone

• Net displacement.

• Distance covered .

Solution :-

As the object is thrown vertically upward then the acceleration due to gravity is taken as negative.

∴ Acceleration due to gravity = -10m/s²

Now using the 3rd equation of motion

→ v² = u² + 2as

Substitute the value we get

→ 0² = 40² + 2×(-10) ×s

→ 0 = 1600 + (-20) ×s

→ -1600 = -20×s

→ s = -1600/-20

→ s = 80m

∴ The maximum height attained by the stone is 80 metre.

Now displacement of the stone

As according to the given statement the stone is thrown vertically upward from initial position and the stone covered maximum height and again the stone attained their initial position . That's why the displacement of the stone is zero.

And

→ Total Distance = 80+80 = 160m

Answer 2

Answer:

80m , Zero and 160m

Explanation:

Applying third equation of motion

=> v^2 = u^2 + 2as

Substitute the value we get

=> 0 = 40^2 + 2×(-10) ×s

=> 0 = 1600 + (-20) ×s

=> -1600 = -20×s

=> s = -1600/-20

=> s = 80m

maximum height is 80m

Now displacement of the stone is 0

=> Total Distance = 80+80 = 160m