Question

Derive the equation for maximum height and range of a projectile

Answer 1

$\huge{\underline{\underline{.........Answer.........}}}$

$\huge{\underline{Given:-}}$

The projectile reaches a height H.

The projectile reaches a height H.and, it covers range R at time t = t.

$\huge{\underline{Explanation:-}}$

Case-1

Maximum height is the height reached by the projectile vertically.

When y = H ; V(y) = 0.

${v}^{2} - {u}^{2} = 2as$

${0}^{2} - {u}^{2} \sin( \theta)^{2} = 2 \times - g \times H \:$

$H = \frac{ {u}^{2} \sin( \theta)^{2}}{2g}$

$\boxed{\boxed{H = \frac{ {u}^{2} \sin( \theta)^{2}}{2g}}}$

Case-2.

It is the maximum horizontal distance travelled by the particle.

x = R. ; t = T.

$x = (u \cos \theta ) \times t$

$x =( u \cos \theta) \times T \:$

But

$T = \frac{2u \sin( \theta) }{g}$

substituting the values.

$R = u \cos \theta \times \frac{2u \sin( \theta) }{g}$

$R = \frac{ {u}^{2}2 \sin( \theta) \cos( \theta) }{g}$

As,

$2 \sin( \theta) \cos( \theta) = \sin(2 \theta)$

substitute the values.

$R = \frac{ {u}^{2} \sin(2 \theta) }{g}$

$\boxed{\boxed{R = \frac{ {u}^{2} \sin(2 \theta) }{g}}}$

Hence equations derived.

# refer the attachment for figure.