a ball is dropped off a 20m high building, the ball weights 5 g. how long does it take to hit the ground and what is the instaneous speed of the the ball at each second of it's fall?
To determine the time it takes for the ball to hit the ground, you can use the equation for motion under constant acceleration:
\[ y = y_0 + v_0t + 0.5at^2 \]
Where:
- \( y \) is the final height (0 meters, since it hits the ground)
- \( y_0 \) is the initial height (20 meters, the height of the building)
- \( v_0 \) is the initial velocity (0 m/s, since the ball is dropped and not thrown)
- \( a \) is the acceleration due to gravity (-9.8 m/s², assuming downward is positive)
- \( t \) is the time we want to find
By substituting the given values into the equation, we get:
\[ 0 = 20 + 0 + 0.5(-9.8)t^2 \]
Simplifying the equation, we have:
\[ 19.6t^2 = 20 \]
Solving for \( t \), we get:
\[ t^2 = \frac{20}{19.6} \]
\[ t = \sqrt{\frac{20}{19.6}} \]
Using a calculator, we find that \( t \approx 1.43 \) seconds. Thus, it takes approximately 1.43 seconds for the ball to hit the ground.
To find the instantaneous speed of the ball at each second of its fall, we can use the equation for instantaneous speed:
\[ v = v_0 + at \]
In this case, the initial velocity \( v_0 \) is 0 m/s, and the acceleration \( a \) is -9.8 m/s². Substituting these values into the equation, we can calculate the speed at each second:
\[ v = 0 + (-9.8)t \]
Let's calculate the instantaneous speed at each second from t = 0 to t = 1.43 seconds:
- At t = 0 seconds, the ball's speed is \( v = 0 + (-9.8) \cdot 0 = 0 \) m/s (initially at rest).
- At t = 1 second, \( v = 0 + (-9.8) \cdot 1 = -9.8 \) m/s (the negative sign indicates downward direction).
- At t = 2 seconds, \( v = 0 + (-9.8) \cdot 2 = -19.6 \) m/s.
- Likewise, at each second, we can calculate the speed by substituting the corresponding value of \( t \) into the equation.
Note: As the ball falls, its speed increases at a constant rate of 9.8 m/s² due to gravity. However, the velocity is negative because it's moving downward.