Question

If the horizontal range and the max height of a body projected at an angle titre to the horizontal is K and Q respectively. Show that the muzzle velocity Vo is given by: Vo=√[2gQ + K²/8Q]​

Answer 1

Step-by-step explanation:

Recall that

`K = (v_0^2\sin2\theta)/(g)\:\:\:\:\:\:\:\:\:(1)`

and

`Q = (v_0^2\sin^2\theta)/(2g)\:\:\:\:\:\:\:\:\:(2)`

From Eqn(2), we can write

`\sin\theta = \sqrt{(2gQ)/(v_0^2)}\:\:\:\:\:\:\:\:\:(3)`

Using the identity
`\sin\theta = 2\sin\theta \cos\theta`, we can rewrite Eqn(1) as

`(gK)/(2v_0^2) = \sin\theta \cos\theta`

Squaring the above equation, we get

`(g^2K^2)/(4v_0^4) = \sin^2\theta \cos^2\theta`

`\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)`

Use Eqn(3) on Eqn(4) and we will get the following:

`(g^2K^2)/(4v_0^4) = (2gQ)/(v_0^2)(1 - (2gQ)/(v_0^2))`

This simplifies to

`(gK^2)/(8v_0^2Q) = 1 - (2gQ)/(v_0^2)`

Rearranging this further, we get

`1 = (2gQ)/(v_0^2) + (gK^2)/(8v_0^2Q)`

Putting
`v_0^2` to the left side, we get

`v_0^2 = 2qQ + (gK^2)/(8Q)`

Finally, taking the square root of the equation above, we get the expression for the muzzle velocity
`v_0` as

`v_0 = \sqrt{2gQ + (gK^2)/(8Q)}`