A flower pot is dropped out of the twentieth story window. How far does the pot drop in 3 seconds? How fast will it be moving after 7 seconds?
vf=g*7sec use 9.8m/s^2 for g.
h= 1/2 g 3^2
To find out how far the flower pot drops in 3 seconds and how fast it will be moving after 7 seconds, we need to consider the equations of motion under constant acceleration.
First, let's calculate the distance dropped in 3 seconds. We'll use the equation:
\[ d = ut + \frac{1}{2} a t^2 \]
where:
- \( d \) is the distance dropped
- \( u \) is the initial velocity (which is 0 since the pot was dropped)
- \( a \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\))
- \( t \) is the time (3 seconds)
Substituting the values into the equation, we have:
\[ d = 0 \cdot 3 + \frac{1}{2} \cdot 9.8 \cdot (3)^2 \]
Simplifying, we get:
\[ d = 0 + \frac{1}{2} \cdot 9.8 \cdot 9 \]
Calculating further:
\[ d = 0 + 44.1 \]
So, the flower pot drops approximately 44.1 meters in 3 seconds.
Next, let's calculate the velocity of the flower pot after 7 seconds. We'll use a different equation of motion:
\[ v = u + at \]
where:
- \( v \) is the final velocity
- \( u \) is the initial velocity (0 m/s)
- \( a \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\))
- \( t \) is the time (7 seconds)
Substituting the values into the equation, we have:
\[ v = 0 + 9.8 \cdot 7 \]
Calculating further:
\[ v = 0 + 68.6 \]
So, the flower pot will be moving at approximately 68.6 m/s after 7 seconds.