A baseball is hit straight up and caught by a catcher 2sec later. The maximum height of the ball during this interval is:
Correct answer: A. 4.9m
- A. 4.9m
- B. 19.6m
- C. 9.8m
- D. 12.6m
Explanation
Given:- Time taken for the baseball to reach its maximum height and return (t) = 2 s- Acceleration due to gravity (g) = 9.8 m/s2Step 1: Time to Reach Maximum Height:The time taken to reach the maximum height is half the total time for the round trip (up and down):Time to reach maximum height = Total time / 2 = 2 s / 2 = 1 sStep 2: Calculation of Maximum Height:Using the equation for displacement in free fall:Displacement = (1/2) x acceleration x time2Displacement = (1/2) 9.8 x (1)2Displacement = (1/2) x 9.8Displacement = 4.9 mThe calculated maximum height reached by the baseball during this interval is 4.9 m, aligning with option (a).
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