A man standing on a hill top projects a stone | vedclass

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(A) The stone is projected horizontally with speed $v_0$ from the origin $(0, 0)$.At any time $t$,the horizontal position is $x = v_0 t$ and the verti

Vedclass · Accepted Answer

$\left(\frac{2 v_0^2 	an 	heta}{g}, -\frac{2 v_0^2 	an ^2 	heta}{g}ight)$

Vedclass · Answer

(A) The stone is projected horizontally with speed $v_0$ from the origin $(0, 0)$.At any time $t$,the horizontal position is $x = v_0 t$ and the vertical position is $y = -\frac{1}{2} g t^2$.The equation of the hill surface passing through the origin with an angle of inclination $	heta$ below the horizontal is $y = -x 	an 	heta$.Substituting the expressions for $x$ and $y$ into the equation of the hill surface:$-\frac{1}{2} g t^2 = -(v_0 t) 	an 	heta$$\frac{1}{2} g t^2 = v_0 t 	an 	heta$Solving for $t$ (where $t 
eq 0$):$t = \frac{2 v_0 	an 	heta}{g}$Now,substitute $t$ back into the expressions for $x$ and $y$:$x = v_0 \left(\frac{2 v_0 	an 	heta}{g}ight) = \frac{2 v_0^2 	an 	heta}{g}$$y = -\frac{1}{2} g \left(\frac{2 v_0 	an 	heta}{g}ight)^2 = -\frac{1}{2} g \left(\frac{4 v_0^2 	an^2 	heta}{g^2}ight) = -\frac{2 v_0^2 	an^2 	heta}{g}$Thus,the coordinates are $\left(\frac{2 v_0^2 	an 	heta}{g}, -\frac{2 v_0^2 	an^2 	heta}{g}ight)$.

$A$ man standing on a hill top projects a stone horizontally with speed $v_0$ as shown in the figure. Taking the coordinate system as given in the figure,the coordinates of the point where the stone will hit the hill surface are:

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