Question

A long jumper leaves at an angle of 30° with the horizontal and at a speed of 18m/s.How far does he jumps

Answer 1

Explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}}\mid}}$

For better understanding refer figure.....

$\huge\bold{GIVEN:-} \\ \\ (\theta) = angle = {30}^{\circ} \\ \\ u = speed \: of \: long \: jumper = 18 m/s \\ \\ For \: finding \: how \: far \: it \: will \: jump = Range \\ \\ \boxed{\boxed{R = \frac{{u}^{2} \sin(2 \theta)}{g} }} \\ \\ R = \frac{18 \times 18 \sin(60)}{10} \\ \\ R = \frac{324 \times \frac{\sqrt{3}}{2}}{10} \\ \\ \huge\boxed{R = 16.2 \times \sqrt{3} m }$

$\boxed{ </p><p>\begin{minipage}{7cm} </p><p></p><p>Some \: formulas \: for \: projectile \\ \\ </p><p> t = time \: of \: flight = \frac{2u \sin(\theta) }{g} \\ \\ H = Max. \: height = \frac{{u}^{2} {\sin}^{2} (\theta)}{2g} \\ \\ R = Range = \frac{{u}^{2} \sin( 2 \theta)}{g} </p><p>\end {minipage}</p><p>}$