Question

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

Answer 1

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: v² = u² + 2as