A rocket is fired upward from the earth's surface such that it creates an acceleration of $19.6 ext{ m/s}^2$

Question

A rocket is fired upward from the earth's surface such that it creates an acceleration of $19.6 	ext{ m/s}^2$. If after $5 	ext{ s}$ its engine is switched off, the maximum height of the rocket from the earth's surface would be:

Accepted Answer

1. **Phase 1: Powered Motion (0 to 5 s):**
   - Initial velocity $u = 0$.
   - Acceleration $a_1 = 19.6 	ext{ m/s}^2$ (upward).
   - Time $t_1 = 5 	ext{ s}$.
   - Height reached ($h_1$): Using $s = ut + \frac{1}{2}at^2$:
     $$ h_1 = 0 + \frac{1}{2}(19.6)(5)^2 = 9.8 	imes 25 = 245 	ext{ m} $$
   - Velocity at engine cut-off ($v_1$): Using $v = u + at$:
     $$ v_1 = 0 + 19.6(5) = 98 	ext{ m/s} $$ .
2. **Phase 2: Free Fall (After 5 s):**
   - The rocket continues to rise due to inertia but is decelerated by gravity.
   - Initial velocity $u_2 = v_1 = 98 	ext{ m/s}$.
   - Acceleration $a_2 = -g = -9.8 	ext{ m/s}^2$.
   - Final velocity at maximum height $v_2 = 0$.
   - Height rose ($h_2$): Using $v^2 = u^2 + 2as$:
     $$ 0^2 = (98)^2 + 2(-9.8)h_2 $$
     $$ 19.6 h_2 = 9604 \implies h_2 = \frac{9604}{19.6} = 490 	ext{ m} $$ .
3. **Total Maximum Height:**
   $$ H_{max} = h_1 + h_2 = 245 	ext{ m} + 490 	ext{ m} = 735 	ext{ m} $$

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