Question

A stone is thrown vertically upwards with an initial velocity of 40m/s. Find the max height reached by the
stone. What is the net displacement and the total distance covered by the stone?​

Answer 1

Given :-

Initial velocity = 40 m/s

Final velocity = 0

To Find :-

The max height reached by the  stone.

The net displacement.

The total distance covered by the stone.

Solution :-

We know that,

• v = Final velocity
• u = Initial velocity
• s = Displacement
• a = Acceleration
• h = Height
• g = Acceleration due to gravity

Using the formula,

$\underline{\boxed{\sf v^2=u^2+2gh}}$

Given that,

Initial velocity (u) = 40 m/s

Final velocity (v) = 0

Acceleration due to gravity (g) = 10 m/s

Substituting their values,

0² = 40² + 2 × 10 × h

h = (40)²/(2×10)

h = 1600/20

h = 80

The stone will reach at the top and reach down.

According to the question,

$\underline{\boxed{\sf Total \ distance=s=h_1+h_2}}$

Given that,

Height (h) = 80 m

Substituting their values,

= 80 + 80

= 160 m

Therefore, the total distance covered by the stone is 160 m.

Then,

$\underline{\boxed{\sf Net \ displacement=h_1-h_2}}$

Given that,

Height (h) = 80 m

Substituting their values,

= 80 - 80

= 0

Therefore, the net displacement is 0.