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?

hf=hi+vi*t-at^2 /2

0=0+6t-9.8t^2/2
t(6-4.9t)=0
time of flight: 6/4.9seconds
max height time is one-half of that.
redo the same equation, knowing time is 1/2 the above t.

To find the maximum height and total time it takes for the stone to come back down, we can use the kinematic equations of motion.

First, let's find the time it takes for the stone to reach its maximum height. We can use the equation:

vf = vi + at

Where:
- vf is the final velocity (0 m/s when the stone reaches its maximum height)
- vi is the initial velocity (6 m/s)
- a is the acceleration (the acceleration due to gravity, -9.8 m/s^2, because the stone is moving upwards against gravity)
- t is the time

Rearranging the equation, we have:

t = (vf - vi) / a

Plugging in the values, we get:

t = (0 m/s - 6 m/s) / (-9.8 m/s^2) = 0.6122 s (approximately)

Now, we need to find the maximum height the stone reaches. We can use the equation:

d = vi * t + (1/2) * a * t^2

Where:
- d is the displacement (the maximum height)
- vi is the initial velocity (6 m/s)
- t is the time (0.6122 s, as found earlier)
- a is the acceleration (-9.8 m/s^2, as it remains the same as in the previous equation)

Plugging in the values, we have:

d = 6 m/s * 0.6122 s + (1/2) * (-9.8 m/s^2) * (0.6122 s)^2 = 1.8366 m (approximately)

Therefore, the maximum height of the stone is approximately 1.8366 meters.

To find the total time it takes for the stone to come back down, we double the time it took to reach the maximum height. So, the total time would be:

Total time = 2 * 0.6122 s = 1.2244 s (approximately)

Therefore, the total time it takes for the stone to come back down is approximately 1.2244 seconds.