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(D) By the law of conservation of energy,the horizontal velocity $v_x$ at height $h_2$ is given by $\frac{1}{2}mv_x^2 = mg(h_1 - h_2)$,so $v_x = \sqrt

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$h_1 = 4h_2$

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(D) By the law of conservation of energy,the horizontal velocity $v_x$ at height $h_2$ is given by $\frac{1}{2}mv_x^2 = mg(h_1 - h_2)$,so $v_x = \sqrt{2g(h_1 - h_2)}$.When the block leaves the slide,it acts as a projectile with initial horizontal velocity $v_x$ and initial vertical velocity $v_{y0} = 0$ from height $h_2$.The vertical velocity $v_y$ just before hitting the ground is given by $v_y^2 = v_{y0}^2 + 2gh_2 = 2gh_2$,so $v_y = \sqrt{2gh_2}$.The angle $	heta$ with the horizontal is given by $	an 	heta = \frac{v_y}{v_x}$.Given $	heta = 30^\circ$,$	an 30^\circ = \frac{1}{\sqrt{3}}$.Thus,$\frac{1}{\sqrt{3}} = \frac{\sqrt{2gh_2}}{\sqrt{2g(h_1 - h_2)}} = \sqrt{\frac{h_2}{h_1 - h_2}}$.Squaring both sides,$\frac{1}{3} = \frac{h_2}{h_1 - h_2}$.$h_1 - h_2 = 3h_2$,which simplifies to $h_1 = 4h_2$.

$A$ block starts from rest at the top of a frictionless slide at a height $h_1$ above the ground. The block leaves the slide moving perfectly horizontally at a height $h_2$ above the ground. The block eventually hits the ground travelling at an angle $\theta = 30^\circ$ below the horizontal. Then:

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