Question:medium

A ball is thrown vertically upwards with a speed of \( 20 \, \text{m/s} \). What is the maximum height reached by the ball? Assume the acceleration due to gravity is \( g = 9.8 \, \text{m/s}^2 \).

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To find the maximum height in vertical motion, use the equation \( v^2 = u^2 + 2as \) where the final velocity \( v = 0 \) at the highest point.
Updated On: Nov 26, 2025
  • \( 20.4 \, \text{m} \)
  • \( 40.8 \, \text{m} \)
  • \( 10.2 \, \text{m} \)
  • \( 50.4 \, \text{m} \)
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The Correct Option is A

Solution and Explanation

Phase 1: Data Assimilation. - Initial velocity: \( u = 20 \, \text{m/s} \) - Terminal velocity at apex: \( v = 0 \, \text{m/s} \) - Gravitational acceleration: \( g = 9.8 \, \text{m/s}^2 \) (vectorially \( a = -9.8 \, \text{m/s}^2 \)) Phase 2: Kinematic Equation Application. Employing the second kinematic equation, which links initial velocity, terminal velocity, acceleration, and vertical displacement (maximum altitude): \[v^2 = u^2 + 2a s\] Inputting the established parameters: \[0 = (20)^2 + 2 \times (-9.8) \times s\] \[0 = 400 - 19.6s\] \[19.6s = 400\] \[s = \frac{400}{19.6} \approx 20.4 \, \text{m}\] Conclusion: The maximum altitude attained by the projectile is approximately \( 20.4 \, \text{m} \).
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