Question

A body of mass m is projected from ground with speed u at an angle theta with horizontal. The power delivered by gravity to it at half of maximum height from ground is???

Answer 1

Consider a projectile of mass `m` launched at a speed `u` at an angle `\theta`
`P(H/2) = (mg)/\sqrt{2}u\sin\theta`

Explanation:

Power: Power is the rate at which energy is delivered-to or drawn-from an object.

`P\equiv (dE)/(dt)`

In this context, the energy delivered-to (downward motion) or drawn-from (upward motion) the projectile is the gravitational potential energy.
Therefore, `P = \frac{dU(y)}{dt}`

If ground is made the zero reference potential, then the gravitational potential energy of a projectile of mass `m` at a height `y` from the ground is : `U(y) = mgy`.

`P(y) = \frac{dU(y)}{dt}=mg\frac{dy}{dt} = mg.v_y(y)` ...... (1)

Thus the power drawn-from/delivered-to the projectile at a given altitude depends on the vertical component of its velocity at that altitude. So we calculate this first.

Consider a projectile motion with an initial launch speed of `v_o` and a launch angle of `\theta_o`. After the launch, the horizontal component of its velocity remains constant. Its vertical component changes with altitude as follows:
`v_y^2(y) = (v_0\sin\theta_0)^2 -2gy`

Therefore the vertical component of the projectile velocity at half the height is then:
`v_y(H/2) = \sqrt{v_o^2\sin^2\theta_o - 2g.H/2} = \sqrt{v_o^2\sin^2\theta_o - gH}`

Height of the projectile is given by: `H = (v_0^2\sin^2\theta_0)/(2g)`

`v_y(H/2) = \sqrt{v_o^2\sin^2\theta_o - \frac{v_0^2\sin^2\theta_0}{2}} = (v_o\sin\theta_0)/\sqrt{2}` ...... (2)

Substituting (2) in (1),
`P(H/2) = (mg)/\sqrt{2}v_o\sin\theta_o`

Given : `v_0=u; \qquad \theta_0 = \theta`
`P(H/2) = (mg)/\sqrt{2}u\sin\theta`