Question

A rocket is fired vertically up from the ground with an acceleration of 10 m/s². The fuel is finished in 1 minute and the rocket continues to move up. The maximum height attained by rocket from ground is

• A. 18 km
• B. 26 km
• C. 36 km
• D. 20 km

Answer 1

Answer: (C)

36 km

Answer 2

The problem describes the motion of a rocket in two phases:

1. Phase 1: Motion with engine thrust (accelerated motion)

   • Initial velocity, $u_1 = 0$ m/s (starts from rest).
   • Acceleration, $a = 10$ m/s².
   • Time duration, $t_1 = 1$ minute = 60 seconds.

   We calculate the height reached ($h_1$) and the velocity at the end of this phase ($v_1$) using kinematic equations:

   • Velocity: $v_1 = u_1 + a t_1$

     $v_1 = 0 + (10 \text{ m/s}^2)(60 \text{ s}) = 600 \text{ m/s}$

   • Height: $h_1 = u_1 t_1 + \frac{1}{2} a t_1^2$

     $h_1 = (0)(60) + \frac{1}{2}(10 \text{ m/s}^2)(60 \text{ s})^2$

     $h_1 = 5 \times 3600 = 18000 \text{ m} = 18 \text{ km}$

2. Phase 2: Motion under gravity (after fuel is finished)

   • The rocket continues to move upwards due to inertia.
   • Initial velocity for this phase, $u_2 = v_1 = 600$ m/s.
   • Acceleration due to gravity, $a_2 = -g = -10$ m/s² (taking upward as positive).
   • Final velocity at maximum height, $v_2 = 0$ m/s.

   We calculate the additional height gained ($h_2$) using the kinematic equation:

   • $v_2^2 = u_2^2 + 2 a_2 h_2$

     $0^2 = (600 \text{ m/s})^2 + 2(-10 \text{ m/s}^2) h_2$

     $0 = 360000 - 20 h_2$

     $20 h_2 = 360000$

     $h_2 = \frac{360000}{20} = 18000 \text{ m} = 18 \text{ km}$

The maximum height attained by the rocket from the ground is the sum of the heights from both phases:

$H_{max} = h_1 + h_2 = 18 \text{ km} + 18 \text{ km} = 36 \text{ km}$