Problem 1:
A ball is dropped from a height of 20 meters. Calculate the time taken for the ball to reach the ground.

Solution:
Given: Initial velocity (u) = 0 m/s (since the ball is dropped)
Final velocity (v) = ? (when the ball hits the ground)
Acceleration (a) = 9.8 m/s^2 (acceleration due to gravity)
Distance (s) = 20 meters
Using the equation of motion: \(v^2 = u^2 + 2as\)
Substitute the values: \(v^2 = 0 + 2 * 9.8 * 20 = 392\)
\(v = \sqrt{392} = 19.8 m/s\)
Now, using the equation: \(v = u + at\), we can find the time taken (t) for the ball to reach the ground.
Substitute the values: \(19.8 = 0 + 9.8t\)
\(t = \frac{19.8}{9.8} = 2 seconds\)
Therefore, it takes 2 seconds for the ball to reach the ground.
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Problem 2:
A stone is thrown vertically downward from a height of 50 meters with an initial velocity of 10 m/s. Calculate the time taken for the stone to hit the ground.

Solution:
Given: Initial velocity (u) = 10 m/s
Final velocity (v) = ? (when the stone hits the ground)
Acceleration (a) = 9.8 m/s^2
Distance (s) = 50 meters
Using the equation of motion: \(v^2 = u^2 + 2as\)
Substitute the values: \(v^2 = 10^2 + 2 * 9.8 * 50 = 960\)
\(v = \sqrt{960} = 30.98 m/s\)
Now, using the equation: \(v = u + at\), we can find the time taken (t) for the stone to hit the ground.
Substitute the values: \(30.98 = 10 + 9.8t\)
\(t = \frac{30.98 - 10}{9.8} = 2.14 seconds\)
Therefore, it takes 2.14 seconds for the stone to hit the ground.
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Problem 3:
An object is released from a height of 30 meters. Calculate the velocity of the object when it hits the ground.

Solution:
Given: Initial velocity (u) = 0 m/s
Final velocity (v) = ? (when the object hits the ground)
Acceleration (a) = 9.8 m/s^2
Distance (s) = 30 meters
Using the equation of motion: \(v^2 = u^2 + 2as\)
Substitute the values: \(v^2 = 0 + 2 * 9.8 * 30 = 588\)
\(v = \sqrt{588} = 24.25 m/s\)
Therefore, the velocity of the object when it hits the ground is 24.25 m/s.