Question

A ball is thrown horizontally from the top of a tower with a speed of $5\text{m/s}$ . What will be the radius of curvature of its trajectory $1\text{s}$ after the ball is thrown? Take $g=10{\text{m/s}}^2$.

• A. $\frac{35}{2}\sqrt{10}\text{m}$
  
• B. $\frac{25}{2}\sqrt{5}\text{m}$
  
• C. $\frac{25}{4}\sqrt{10}\text{m}$
  
• D. $\frac{25}{4}\sqrt{5}\text{m}$
  

Answer 1

The correct option is B $\frac{25}{2}\sqrt{5}\text{m}$


For the given horizontal projection:
$u_x=5\text{m/s},u_y=0,a_x=0,a_y=+g$
Reference Y-axis has been taken vertically downwards as +ve.

Horizontal velocity does not change. Hence velocities after $t=1\text{s}$ are
$v_{x_1}=5\text{m/s}{v}_{y_1}=u_y+at=0+10\times 1=10\text{m/s}$

Magnitude of velocity after $t=1\text{s}$ is:
$v_1=\sqrt{v_{x_1}^2+v_{y_1}^2}=\sqrt{5^2+{10}^2}=\sqrt{25+100}=\sqrt{125}=5\sqrt{5}\text{m/s}$


If you consider a circle as shown in figure with $\overrightarrow{v_1}$ tangential, magnitude of radial acceleration of ball,
$a_r=g\cos \theta$

From the shown figure of vector triangle ;
$\cos \theta =\frac{v_{x_1}}{v_1}=\frac{5}{5\sqrt{5}}$
Therefore,
$a_r=10\times \frac{5}{5\sqrt{5}}=2\sqrt{5}{\text{m/s}}^2$
$\therefore$ Radius of curvature
$r=\frac{v_1^2}{a_r}=\frac{125}{2\sqrt{5}}=\frac{25}{2}\sqrt{5}\text{m}$