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A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate the maximum height to which it rises,
1. 122.5 m
2. 124.5 m
3. 126.5 m
4. 128.5 m

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Correct Answer - Option 1 : 122.5 m

The correct answer is 122.5 m

  • Given data
    • Initial velocity U = 49 m/s
    • Final speed v at maximum height = 0
    • Acceleration due to earth gravity g = -9.8 m/s2 (thus negative as the ball is thrown up).
    • By third equation of motion, V2 = U2 – 2gS 
    • Substitute all the values in the above equation, 0 = 492 - 2 × 9.8 × S
    • S = 492 / (2 × 9.8)
    • S = 122.5 m 

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  • There are three equations of motion that can be used to derive components such as displacement(s), velocity (initial and final), time(t) and acceleration(a).
  • The following are the three equation of motion:
    • First Equation of Motion: v = u + at
    • Second Equation of Motion: s = ut + 1/2(at2)
    • Third Equation of Motion: v= u+ 2as

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