Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

245 m

490 m

980 m

735 m

Step-by-Step Solution

Phase 1: Powered Motion (0 to 5 s):

Initial velocity $u = 0$ .
Acceleration $a_1 = 19.6 \text{ m/s}^2$ (upward).
Time $t_1 = 5 \text{ s}$ .
Height reached ( $h_1$ ): Using $s = ut + \frac{1}{2}at^2$ : $h_1 = 0 + \frac{1}{2}(19.6)(5)^2 = 9.8 \times 25 = 245 \text{ m}$
Velocity at engine cut-off ( $v_1$ ): Using $v = u + at$ : $v_1 = 0 + 19.6(5) = 98 \text{ m/s}$ .

Phase 2: Free Fall (After 5 s):

The rocket continues to rise due to inertia but is decelerated by gravity.
Initial velocity $u_2 = v_1 = 98 \text{ m/s}$ .
Acceleration $a_2 = -g = -9.8 \text{ m/s}^2$ .
Final velocity at maximum height $v_2 = 0$ .
Height rose ( $h_2$ ): Using $v^2 = u^2 + 2as$ : $0^2 = (98)^2 + 2(-9.8)h_2$ $19.6 h_2 = 9604 \implies h_2 = \frac{9604}{19.6} = 490 \text{ m}$ .

Total Maximum Height: $H_{max} = h_1 + h_2 = 245 \text{ m} + 490 \text{ m} = 735 \text{ m}$

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