A stone is thrown vertically upward with a velocity of 40 m/s. Taking g = 10 m/s2, find themaximum height reached by the stone. What is the net displacement and the total distancecovered by the stone?

Step 1: Identify the given information:

• Initial velocity (u) = 40 m/s
• Acceleration due to gravity (g) = 10 m/s^2

Step 2: Determine the maximum height reached by the stone: To find the maximum height reached by the stone, we can use the kinematic equation for vertical motion:

v^2 = u^2 + 2as

Where:

• v is the final velocity (which is 0 m/s at the maximum height)
• u is the initial velocity (40 m/s)
• a is the acceleration due to gravity (-10 m/s^2, considering the upward direction as positive)
• s is the displacement (maximum height)

Rearranging the equation, we have:

0^2 = (40 m/s)^2 + 2(-10 m/s^2)s

0 = 1600 m^2/s^2 - 20s

20s = 1600 m^2/s^2

s = 1600 m^2/s^2 / 20

s = 80 m

Therefore, the maximum height reached by the stone is 80 meters.

Step 3: Calculate the net displacement: The net displacement is the difference between the final and initial positions of the stone. Since the stone is thrown vertically upward and comes back down, the net displacement is zero. This is because the stone ends up at the same position from where it was thrown.

Therefore, the net displacement is 0 meters.

Step 4: Determine the total distance covered by the stone: The total distance covered by the stone is the sum of the distances traveled during the upward and downward motion.

During the upward motion, the stone covers a distance equal to the maximum height reached, which is 80 meters.

During the downward motion, the stone covers the same distance as the upward motion, which is also 80 meters.

Therefore, the total distance covered by the stone is 80 meters + 80 meters = 160 meters.