Question

An object is thrown horizontally with a velocity of $10\text{m/s}$. Find the radius of curvature of its trajectory $3\text{s}$ after the motion has begun. Take $g=10{\text{m/s}}^2$

• A. $100\sqrt{10}\text{m}$
• B. $316\text{m}$
• C. $100\sqrt{5}\text{m}$
• D. $300\text{m}$

Answer 1

The correct option is A $100\sqrt{10}\text{m}$

For the given horizontal projection:
$u_x=10\text{m/s},u_y=0,a_x=0,a_y=+g$
(taking downwards as +ve)

Horizontal velocity does not change. Hence velocities after $t=3\text{s}$ are,
$v_{x_3}=10\text{m/s}{v}_{y_3}=u_y+at=0+10\times 3=30\text{m/s}$

Magnitude of velocity after $t=3\text{s}$ is:
$v_3=\sqrt{v_{x_3}^2+v_{y_3}^2}=\sqrt{{10}^2+{30}^2}=\sqrt{100+900}=\sqrt{1000}=10\sqrt{10}\text{m/s}$


Magnitude of radial acceleration of ball at the point where $\overrightarrow{v_3}$ serves as tangential velocity is
$a_r=g\cos \theta$

In triangle $ABC$,
$\cos \theta =\frac{AB}{AC}=\frac{v_{x_3}}{v_3}=\frac{10}{10\sqrt{10}}=\frac{1}{\sqrt{10}}$
Therefore, $a_r=10\times \frac{1}{\sqrt{10}}=\sqrt{10}{\text{m/s}}^2$
and Radius of curvature
$r=\frac{v_3^2}{a_r}=\frac{1000}{\sqrt{10}}=100\sqrt{10}\text{m}$