While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.87 m/s. The stone subsequently falls to the ground, which is 13.5 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? How much time is the stone in the air? Ignore air resistance and take g = 9.81 m/s2.

Final KE+final PE=initialKE + initial PE

1/2 m vf^2+0=1/2 m (6.87)^2 + mg(13.5)

solve for vf

time in air:
vtop=0=vi*t-g t^2
o=6.87t-9.81 t^2
solve for t.

To find the speed of the stone when it impacts the ground and the time it is in the air, we can use the kinematic equations of motion.

Let's break down the problem step by step.

Step 1: Determine the time taken to reach the highest point:

When the stone reaches its highest point, its vertical velocity becomes zero. We can use the equation:

v = u + gt

where:
v = final velocity (zero at the highest point)
u = initial velocity (6.87 m/s)
g = acceleration due to gravity (-9.81 m/s^2)
t = time taken

Rearranging the equation to solve for time (t):

0 = 6.87 - 9.81t

9.81t = 6.87

t = 6.87 / 9.81

t = 0.7 seconds (approximately)

Step 2: Determine the time taken to fall from the highest point to the ground:

Since the stone takes the same amount of time to reach the highest point as it takes to fall from the highest point to the ground, the total time in the air is:

2 * 0.7 seconds = 1.4 seconds

Step 3: Determine the speed at impact:

Using the equation:

v = u + gt

where:
v = final velocity (unknown)
u = initial velocity (6.87 m/s)
g = acceleration due to gravity (9.81 m/s^2)
t = time taken (1.4 seconds)

v = 6.87 + (9.81 * 1.4)

v ≈ 19.6 m/s

Therefore, the speed at which the stone impacts the ground is approximately 19.6 m/s and the stone is in the air for approximately 1.4 seconds.