Question

An object of mass 'm' is projected with an initial velocity 'u' with an angle of '\(\theta\)' with the horizontal. The average power delivered by gravity in reaching the highest point is

• A. \(\frac{mgu \sin^2 \theta}{2}\)
• B. \(\frac{mu^2 \sin^2 \theta}{2g}\)
• C. \(\frac{mg \sin \theta}{2u}\)
• D. \(\frac{mgu \sin \theta}{2}\)

Hint:

When dealing with projectile motion, breaking down the motion into horizontal and vertical components simplifies analysis. Power calculations often rely on understanding how energy changes over time.

Answer 1

The Correct Option is D

The object reaches its highest point when the vertical component of its velocity becomes zero. 
The initial vertical velocity is \(u \sin \theta\), and it takes time \(\frac{u \sin \theta}{g}\) to reach this point due to gravity. 
The change in gravitational potential energy, which is equal to the work done by gravity, is: \[ \Delta U = mgh = mg \left(u \sin \theta \cdot \frac{u \sin \theta}{g}\right) = mu^2 \sin^2 \theta \] The average power delivered by gravity is the work done by gravity divided by the time taken to reach the highest point: \[ P_{avg} = \frac{\Delta U}{t} = \frac{mu^2 \sin^2 \theta}{\frac{2u \sin \theta}{g}} = \frac{mgu \sin \theta}{2} \]