Question:

A cannon ball is fired from the top of a \(55\,\text{m}\) high cliff with an initial speed of \(50\,\text{m s}^{-1}\). The speed of the cannon ball while hitting the ground in \(\text{m s}^{-1}\) is \((\text{acceleration due to gravity}=10\,\text{m s}^{-2})\)

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When only the speed at a lower height is required and air resistance is neglected, use \[ v^2=u^2+2gh. \] This result is independent of the direction of initial projection.
Updated On: Jun 18, 2026
  • \(50\)
  • \(60\)
  • \(33.2\)
  • \(83.2\)
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The Correct Option is B

Solution and Explanation

Step 1: Use conservation of mechanical energy.
The cannon ball is fired from a height of \[ h=55\,\text{m} \] with initial speed \[ u=50\,\text{m s}^{-1}. \] When it reaches the ground, its gravitational potential energy is converted into additional kinetic energy.

Step 2: Write the energy relation.

Using \[ v^2=u^2+2gh, \] where \(v\) is the speed while hitting the ground.
Substituting the values, \[ v^2=50^2+2(10)(55) \] \[ v^2=2500+1100 \] \[ v^2=3600 \]

Step 3: Find \(v\).

\[ v=\sqrt{3600} \] \[ v=60\,\text{m s}^{-1} \]

Step 4: Final conclusion.

Therefore, the speed of the cannon ball while hitting the ground is \[ \boxed{60\,\text{m s}^{-1}} \]
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