while standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 7.79 m/s. the stone then falls to the ground 14.3 m below. At what speed does the stone impact the ground? how much time is it in the air? g=9.81 m/s^2 and ignore air resistance
To determine the speed at which the stone impacts the ground and the time it stays in the air, we need to break down the problem into two parts: the stone's upward motion and its downward motion.
Let's start by calculating the time it takes for the stone to reach its maximum height during the upward motion. We can use the kinematic equation:
v = u + at
Where:
v = final velocity (0 m/s since it reaches its maximum height)
u = initial velocity (7.79 m/s)
a = acceleration due to gravity (-9.81 m/s^2)
t = time
Rearranging the equation to solve for time (t):
t = (v - u) / a
t = (0 - 7.79) / -9.81
t ≈ 0.794 seconds
Now, we know that it takes approximately 0.794 seconds to reach its maximum height. Since the stone is at rest at its maximum height, it will take the same amount of time to fall back down. So, the total time of flight will be twice the time to reach the maximum height.
Total time of flight = 2 * 0.794
Total time of flight ≈ 1.588 seconds
Next, we'll calculate the stone's final velocity when it impacts the ground during the downward motion. To do this, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity (0 m/s since it starts at rest at the maximum height)
a = acceleration due to gravity (-9.81 m/s^2)
s = distance (14.3 m)
Rearranging the equation to solve for final velocity (v):
v = √(u^2 + 2as)
v = √(0 + 2 * -9.81 * -14.3)
v ≈ √(0 + 278.9)
v ≈ √278.9
v ≈ 16.7 m/s (rounded to one decimal place)
Therefore, the stone impacts the ground with a speed of approximately 16.7 m/s, and it remains in the air for approximately 1.588 seconds.