14. You throw a 2 kg. Stone straight up into the air with a velocity of 6m/s. What is the max height of the stone and the total time it takes for it to come back down?

Time in the air = 2 Vo/g

Max height = Vo^2/(2g)

Vo is the initial velocity when thrown

To find the maximum height and the total time it takes for the stone to come back down, we can break down the problem using the principles of projectile motion.

First, let's analyze the vertical motion of the stone. We can use the kinematic equation for vertical displacement:

h = (v^2 - u^2) / (2g)

Where:
h = maximum height
v = final velocity (0 m/s when the stone reaches its highest point)
u = initial velocity (6 m/s)
g = acceleration due to gravity (9.8 m/s^2)

Substituting the given values into the equation:
h = (0^2 - 6^2) / (2 × 9.8)
h = (-36) / 19.6
h ≈ -1.84 meters

Since height cannot be negative, we discard the negative sign and conclude that the maximum height of the stone is approximately 1.84 meters.

Next, to find the total time it takes for the stone to come back down, we can use the equation for time of flight:

t = 2u / g

Substituting the given values into the equation:
t = 2 × 6 / 9.8
t ≈ 1.22 seconds

Therefore, the total time it takes for the stone to come back down is approximately 1.22 seconds.