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A rocket is fired upward from the earth's surface such that it creates an acceleration of 19.6 m/sec2. If after 5 sec its engine is switched off, the maximum height of the rocket from the earth's surface would be
  • a)
    245 m
  • b)
    490 m
  • c)
    980 m
  • d)
    735 m
Correct answer is option 'D'. Can you explain this answer?
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Given: Acceleration of rocket = 19.6 m/sec2, Time taken to switch off engine = 5 sec

To find: Maximum height of rocket from the earth's surface

We can use the kinematic equation:

v^2 = u^2 + 2as

where, v = final velocity (when engine is switched off), u = initial velocity (when rocket is fired), a = acceleration, s = distance travelled

Initial velocity, u = 0 (since the rocket was fired from rest)

Final velocity, v = u + at = 0 + 19.6 x 5 = 98 m/sec

Distance travelled, s = ut + 1/2at^2 = 0 x 5 + 1/2 x 19.6 x 5^2 = 245 m

Now, we can use the formula for maximum height reached by the rocket:

h = s + u^2/2g

where, g = acceleration due to gravity = 9.8 m/sec2

Substituting the values, we get:

h = 245 + 0^2/2 x 9.8 = 245 + 0 = 245 m

Therefore, the maximum height of the rocket from the earth's surface is 245 m.
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