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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Plane

From a point on the group at a distance 2 meters from the foot of a vertical wall , a ball is thrown at an angle $45^{\circ}$ which just clear the top of the wall and afterwards strikes the ground at a distance 4 m on the other side. The height of the wall is

$\begin{array}{1 1}(A)\; \frac{1}{3} \;m\\(B)\;\frac{2}{3}\;m \\(C)\;\frac{3}{4} \;m\\(D)\;\frac{4}{3}\;m \end{array} $

1 Answer

Using the equation of trajectory and applying for the point $B( 2,h)$ and also for the point $C(6,0)$ we get,
$h= 2 \tan 45^{\circ}- \large\frac{g(2)^2}{2 v_0 ^2 \cos ^2 45^{\circ}}$------(i)
$0= 6 \tan 45^{\circ}$$ - \large\frac{g(6)^2}{2v_0^2\cos ^2 45^{\circ}}$-----(ii)
Putting the value of $\large\frac{g}{2v_90^2 \cos ^2 45^{\circ}}$ from (ii) and (i) we
$h= 2 - \large\frac{1}{6}$$(2)^2 =\large\frac{4}{3}$$\;m$
Hence D is the correct answer.
answered Jul 14, 2014 by meena.p
 

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