A 1 kg object is dropped from a height. What is its velocity just before it hits the ground (neglect air resistance)? (2020)

1 question

1 item

Q: A 1 kg object is dropped from a height. What is its velocity just before it hits the ground (neglect air resistance)? (2020)

Solution: Using v² = u² + 2gh, v = √(0 + 2 × 9.8 m/s² × h). For h = 20 m, v = √(392) = 20 m/s.

Steps: 8

Step 1: Identify the initial velocity (u) of the object. Since it is dropped, u = 0 m/s.
Step 2: Identify the acceleration due to gravity (g). This is approximately 9.8 m/s².
Step 3: Identify the height (h) from which the object is dropped. In this case, h = 20 m.
Step 4: Use the formula v² = u² + 2gh to find the final velocity (v) just before hitting the ground.
Step 5: Substitute the values into the formula: v² = 0 + 2 × 9.8 m/s² × 20 m.
Step 6: Calculate the right side: 2 × 9.8 × 20 = 392.
Step 7: Take the square root of 392 to find v: v = √392.
Step 8: Calculate √392, which is approximately 20 m/s.