Question

Astone has thrown vehically upward with and inicial velocity of 40m/s taking g=10m/s2 fine the maximum height reached by the stone.What is the net displacement and the total distance convered by the stone​

Answer 1

Answer

• Maximum height = 80 m
• Distance = 160 m
• Displacement = 0 m

Explanation

Given

• We are throwing a stone vertically upward
• Acceleration = -10 m/s²
• Initial velocity = 40 m/s

To Find

• Maximum height attained?
• Distance = ?
• Displacement = ?

Solution

We shall use the third equation of motion to find the maximum height reached and then with that we shall find the distance and the displacement of the body

Maximum Height

➝ v²-u² = 2as

➝ 0²-40² = 2 × (-10) × s

➝ 0-1600 = -20s

➝ -1600 = -20s

➝ 1600 = 20s

[Cancel -ve sign]

➝ 1600/20 = s

➝ Distance = 80 m

Distance Travelled

• The stone will once go up and then come down once so which means it has to travel the same distance twice

➝ Total Distance = 80+80

➝ Total Distance = 160 m

Displacement of the stone

• Displacement is the shortest distance between two points so here the ball will start and end at the same point

➝ Displacement = 0 m