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

Question

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

Accepted Answer

1. **Identify the Given Data:**
   - Initial velocity, $u = 19.6 	ext{ m/s}$.
   - Final velocity at maximum height, $v = 0 	ext{ m/s}$.
   - Acceleration due to gravity, $a = -g = -9.8 	ext{ m/s}^2$ (acting downwards).
2. **Apply Kinematic Equation:** Use the equation connecting velocity, acceleration, and displacement: $v^2 = u^2 + 2as$ .
   - Here, displacement $s = h$ (maximum height).
   $$ 0^2 = (19.6)^2 + 2(-9.8)h $$
   $$ 0 = (19.6)^2 - 19.6h $$
   $$ 19.6h = (19.6)^2 $$
3. **Solve for Height ($h$):**
   $$ h = \frac{19.6 	imes 19.6}{19.6} = 19.6 	ext{ m} $$
   Alternatively, using the maximum height formula $h_m = \frac{u^2}{2g}$ :
   $$ h_m = \frac{(19.6)^2}{2 	imes 9.8} = \frac{19.6 	imes 19.6}{19.6} = 19.6 	ext{ m} $$

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