A ball is projected horizontally with a speed | vedclass

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Question

(B) Let the ball be projected from the origin $(0,0)$. The equation of the trajectory for horizontal projection is $y = \frac{1}{2} g t^2$ and $x = vt

Vedclass · Accepted Answer

$\sqrt{2} \left[ \frac{2v^2}{g} \right]$

Vedclass · Answer

(B) Let the ball be projected from the origin $(0,0)$. The equation of the trajectory for horizontal projection is $y = \frac{1}{2} g t^2$ and $x = vt$.Substituting $t = \frac{x}{v}$ into the equation for $y$,we get $y = \frac{1}{2} g \left( \frac{x}{v} ight)^2 = \frac{gx^2}{2v^2}$.The inclined plane makes an angle of $45^{\circ}$ with the horizontal. The equation of the plane is $y = x 	an(45^{\circ}) = x$.Equating the two expressions for $y$,we have $x = \frac{gx^2}{2v^2}$.Solving for $x$,we get $x = \frac{2v^2}{g}$.Since $y = x$,the vertical coordinate is also $y = \frac{2v^2}{g}$.The distance $l$ from the point of projection is given by $l = \sqrt{x^2 + y^2}$.Substituting the values of $x$ and $y$,$l = \sqrt{x^2 + x^2} = x\sqrt{2}$.Therefore,$l = \sqrt{2} \left[ \frac{2v^2}{g} ight]$.

$A$ ball is projected horizontally with a speed $v$ from the top of an inclined plane which is inclined at an angle $45^{\circ}$ with the horizontal. How far from the point of projection will the ball strike the plane?

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