A flower pot falls from a windsill 30 meters above the sidewalk. How fast is it going when it hits the ground? How long does it take to reach the ground? How fast would it be going if it was thrown down at 8m/s instead?
vf^2=vi^2+2gd
for how long, vf=gt
To determine the speed and time it takes for the flower pot to hit the ground, we can use the kinematic equations of motion. The key equation we'll use is:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
1. Speed of the falling flower pot:
Since the flower pot falls freely due to gravity, its initial velocity (u) is 0 m/s (as it's dropped from rest with no initial downward velocity). The acceleration (a) due to gravity is approximately 9.8 m/s^2, considering Earth's gravity.
Plugging in the values into the equation:
v^2 = 0^2 + 2 * 9.8 * 30,
v^2 = 0 + 588,
v^2 = 588,
v ≈ √588 ≈ 24.25 m/s.
Therefore, the flower pot is moving at approximately 24.25 m/s when it hits the ground.
2. Time taken to reach the ground:
To find the time (t) it takes for the flower pot to hit the ground, we can use the equation:
s = ut + 0.5 * a * t^2.
In this case, s is the displacement (30 m), u is the initial velocity (0 m/s), and a is the acceleration due to gravity (-9.8 m/s^2).
Plugging in the values:
30 = 0 * t + 0.5 * (-9.8) * t^2,
30 = -4.9 * t^2,
6t^2 = -30,
t^2 = -30 / -6,
t^2 = 5,
t ≈ √5 ≈ 2.24 s.
Therefore, it takes approximately 2.24 seconds for the flower pot to reach the ground.
3. Speed if thrown down at 8 m/s:
If the flower pot is thrown down with an initial velocity (u) of 8 m/s, we can use the same equation to calculate the final velocity (v).
Plugging in the values and solving for v:
v^2 = 8^2 + 2 * 9.8 * 30,
v^2 = 64 + 588,
v^2 = 652,
v ≈ √652 ≈ 25.51 m/s.
Thus, if the flower pot is thrown down with an initial velocity of 8 m/s, it would be moving at approximately 25.51 m/s when it hits the ground.