Question

A particle is projected in air from origin with speed u and at an angle $\theta$ with positive x-axis At its maximum height its trajectory has radius of curvature $R$ and center of curvature has co-ordinate $(20m,\frac{70}{3}m)$ choose the correct option(s) $(g=-10m/s^2)$

• A. Equation of trajectory of the projectile motion is $y=3x-\frac{3x^2}{40}$
• B. The maximum height of the projectile is 30 m
• C. The value of u is $\sqrt{\frac{2000}{3}}$ m/s
• D. The value of $\theta$ is ${\tan }^{-1}\left(\frac{3}{\sqrt{10}}\right)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Equation of trajectory of the projectile motion is $y=3x-\frac{3x^2}{40}$
B The maximum height of the projectile is 30 m
C The value of u is $\sqrt{\frac{2000}{3}}$ m/s

Range = 40 m = $\frac{u^2\sin 2\theta }{g}\Rightarrow u^2\sin 2\theta =400$

$R=\frac{u^2{\cos }^2\theta }{g}\Rightarrow$ radius of curvature at maximum height $H=\frac{u^2{\sin }^2\theta }{2g}$

$4\times u^4{\sin }^2\theta {\cos }^2\theta =(40C{)}^2\Rightarrow 4\times Rg\times H\times 2g=400\times 400$

$\Rightarrow HR=\frac{400\times 400}{4\times 10\times 2\times 10}=\mathrm{200...........}\left(i\right)$

$H-R=\frac{70}{3}...................\left(ii\right)$

$(H-R{)}^2=(H-R{)}^2+4HR=\frac{4900}{9}+800$

$H+R=\frac{110}{3}.............\left(iii\right)$

with the help of equation H = 30 m, R = 20/3 m

$\frac{H}{Range}=\frac{\tan \theta }{4}=\frac{30}{40}\Rightarrow \tan \theta =3\Rightarrow \sin \theta =-\frac{3}{\sqrt{10}}$

$u^2=\frac{2gH}{\sin \theta }=\frac{2\times 10\times 30\times \sqrt{10}\times \sqrt{10}}{3\times 3}\Rightarrow u=\sqrt{\frac{2000}{3}}m/s$