An object is thrown horizontally from a height of 45 m. How long does it take to hit the ground? (g = 9.8 m/s²)

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Q: An object is thrown horizontally from a height of 45 m. How long does it take to hit the ground? (g = 9.8 m/s²)

Solution: Using the formula s = 0.5gt², we have 45 = 0.5 * 9.8 * t², solving gives t ≈ 3.03 s.

Steps: 9

Step 1: Identify the height from which the object is thrown. In this case, the height is 45 meters.
Step 2: Understand that the only force acting on the object in the vertical direction is gravity, which accelerates the object downwards at 9.8 m/s².
Step 3: Use the formula for distance fallen under constant acceleration: s = 0.5 * g * t², where 's' is the distance (45 m), 'g' is the acceleration due to gravity (9.8 m/s²), and 't' is the time in seconds.
Step 4: Substitute the known values into the formula: 45 = 0.5 * 9.8 * t².
Step 5: Simplify the equation: 45 = 4.9 * t² (since 0.5 * 9.8 = 4.9).
Step 6: To isolate t², divide both sides by 4.9: t² = 45 / 4.9.
Step 7: Calculate the right side: t² ≈ 9.18.
Step 8: Take the square root of both sides to find t: t ≈ √9.18.
Step 9: Calculate the square root: t ≈ 3.03 seconds.