Question

a projectile just goes over a wall of height h metre and at a distance d metre apart from the point of projectile and later it hits a amrk at a height h metre and distance 2d metre show that the velocity of projectile v is given by - 4v^2\g= 4d^2+9h^2\h

Answer 1

✪ᴀɴsᴡᴇʀ✪

☆ɢɪᴠᴇɴ☆

• A projectile pass a point at height 'h' which is at distance 'd' from point of projection.

• The same projectile pass another point at height 'h' which is at distance '2d' from point of projection.

☆ᴛᴏ ᴘʀᴏᴠᴇ☆

• Velocity of projection is given by

$\:\:\:\:\:\:\:\:\:\:\:\: \large{\frac{4u^2}{g} = \frac{4d^2 + 9h^2}{h}}$

᯽ғᴏʀᴍᴜʟᴀ᯽

• Equation of trajectory of a projectile is given by

$\:\:\:\:\boxed{\green{y = [tan(\theta)]x - \left[\frac{g}{2u^2cos^2(\theta)} \right]x^2}}$

☆ᴘʀᴏᴏғ☆

Since the particle pass through (d, h), using the trajectory equation,

$\to h = [tan(\theta)]d - \left[\frac{g}{2u^2cos^2(\theta)} \right]d^2$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-----(1)$

Also, the particle pass through (2d, h) so

$\to h = 2[tan(\theta)]d - 4\left[\frac{g}{2u^2cos^2(\theta)} \right]d^2$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-----(2)$

Multiplying (1) by 2 and substracting (2), we get

$\to h = \left[\frac{(-2+4)g}{2u^2cos^2(\theta)}\right]d^2$

$\to h = \left[\frac{(-2+4)g}{2u^2cos^2(\theta)}\right]d^2$

$\to h = \left[\frac{2gd^2}{2u^2cos^2(\theta)}\right]$

$\to cos^2(\theta) = \left[\frac{gd^2}{u^2h}\right]$

$\to cos(\theta) =\sqrt{\left[\frac{gd^2}{u^2h}\right]}$

$\to tan(\theta) =\frac{\sqrt{u^2h - d^2g}} {\sqrt{d^2g} }$

Substituting these values in (1), we get

$\to h = \frac{\sqrt{u^2h - d^2g}} {\sqrt{d^2g}}d - \left[\frac{g}{2u^2\left(\frac{gd^2}{u^2h} \right)} \right]d^2$

$\to h = \frac{\sqrt{u^2h - d^2g}} {d\sqrt{g}}d - \frac{h}{2}$

$\to h+\frac{h}{2}= \frac{\sqrt{u^2h - d^2g}} {\sqrt{g}}$

$\to (\frac{3h}{2})^2 = \frac{u^2h}{g} - d^2$

$\to (\frac{3h}{2})^2 + d^2 = \frac{u^2h}{g}$

$\to 9h^2 + 4d^2 = 4\frac{u^2h}{g}$

$\to \boxed{\blue{\frac{4u^2}{g} = \frac{4d^2 + 9h^2}{h}}}$