Question

a small ball travels with speed v0 over a fixed wedge as shown in the figure the minimum value of theta so that the particle does not hit the inclined plane will be​

Answer 1

Given:

A small ball travels with speed v0 over a fixed wedge as shown in the figure.

To find:

Minimum value of $\theta$ for which the ball will not touch the inclined plane ?

Diagram:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.4mm}\qbezier(0,5)(5,4.5)(5,0)\qbezier(0,5)(0,5)(5,0)\put(0,0){\line(0,1){5}}\put(-2,0){\line(1,0){8}}\put(0,5){\vector(1,0){3.5}}\multiput(-0.5,0)(0.5,0){13}{\qbezier(0,0)(0,0)(-0.3,-0.3)}\put(-1,2.5){\bf{H}}\put(4,4.9){ \bf{u_x= v_{0}} }\put(2.3,-1){\bf{R}}\qbezier(4.3,0)(4.2,0.3)(4.5,0.5)\put(3.5,0.3){ \Theta }\put(0,5){\circle*{0.5}}\put(0,5){\line(-1,0){2}}\end{picture}$

Calculation:

The minimum value for $\theta$ have to such that the ball after being projected from the top directly lands at the bottom point of wedge.

(Refer to attached diagram).

Let time taken to reach bottom be t :

$\therefore \: H = u_{y}t + \dfrac{1}{2} g {t}^{2}$

$\implies\: H = (0)t + \dfrac{1}{2} g {t}^{2}$

$\implies\: H = \dfrac{1}{2} g {t}^{2}$

$\implies\: {t}^{2} = \dfrac{2H}{g}$

$\implies\: t = \sqrt{ \dfrac{2H}{g} }$

Now , range of the projectile:

$\therefore \: R = u_{x} \times t$

$\implies \: R = v_{0}\times \sqrt{ \dfrac{2H}{g} }$

Now , we can say that:

$\therefore \: \tan( \theta) = \dfrac{H}{R}$

$\implies\: \tan( \theta) = \dfrac{H}{ v_{0} \sqrt{ \frac{2H}{g} } }$

$\implies\: \tan( \theta) = \dfrac{H}{ v_{0} } \sqrt{ \dfrac{g}{2H} }$

$\implies\: \tan( \theta) = \dfrac{1}{ v_{0} } \sqrt{ \dfrac{g {H}^{2} }{2H} }$

$\implies\: \tan( \theta) = \dfrac{1}{ v_{0} } \sqrt{ \dfrac{g H}{2} }$

$\implies\: \theta = { \tan}^{ - 1} \bigg( \dfrac{1}{ v_{0} } \sqrt{ \dfrac{g H}{2} } \bigg )$

So, final answer is:

$\boxed{ \bold{\: \theta = { \tan}^{ - 1} \bigg( \dfrac{1}{ v_{0} } \sqrt{ \dfrac{g H}{2} } \bigg )}}$