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A cricket ball is thrown up with a speed of 19.6 ms−119.6 \text{ ms}^{-1}. The maximum height it can reach is:

A

9.8 m

B

19.6 m

C

29.4 m

D

39.2 m

Step-by-Step Solution

  1. Identify the Given Data:
  • Initial velocity, u=19.6 m/su = 19.6 \text{ m/s}.
  • Final velocity at maximum height, v=0 m/sv = 0 \text{ m/s}.
  • Acceleration due to gravity, a=−g=−9.8 m/s2a = -g = -9.8 \text{ m/s}^2 (acting downwards).
  1. Apply Kinematic Equation: Use the equation connecting velocity, acceleration, and displacement: v2=u2+2asv^2 = u^2 + 2as .
  • Here, displacement s=hs = h (maximum height). 02=(19.6)2+2(−9.8)h0^2 = (19.6)^2 + 2(-9.8)h 0=(19.6)2−19.6h0 = (19.6)^2 - 19.6h 19.6h=(19.6)219.6h = (19.6)^2
  1. Solve for Height (hh): h=19.6×19.619.6=19.6 mh = \frac{19.6 \times 19.6}{19.6} = 19.6 \text{ m} Alternatively, using the maximum height formula hm=u22gh_m = \frac{u^2}{2g} : hm=(19.6)22×9.8=19.6×19.619.6=19.6 mh_m = \frac{(19.6)^2}{2 \times 9.8} = \frac{19.6 \times 19.6}{19.6} = 19.6 \text{ m}
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