本题目来源于试卷: 2012美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A cannonball is launched +ilo,paq-p+z,tlz v+ with initial velocity of mxo 4gwy( l3nxa 9(+1izeevz6 lagnitude $v_{0}$ over a horizontal surface. At what minimum angle $\theta_{min}$ above the horizontal should the cannonball be launched so that it rises to a height $H$ which is larger than the horizontal distance $R$ that it will travel when it returns to the ground?
A. $\theta_{min} = 76^\circ$
B. $\theta_{min} = 72^\circ$
C. $\theta_{min} = 60^\circ$
D. $\theta_{min} = 45^\circ$
E. There is no such angle, as $R > H$ for all range problems.
参考答案: A
本题详细解析:
The maximum height is lh-o l+ p9eqkk3d.ba 7$H = \frac{(v_0 \sin \theta)^2}{2g}$. The horizontal range is $R = \frac{v_0^2 \sin(2\theta)}{g} = \frac{2 v_0^2 \sin \theta \cos \theta}{g}$. We require $H > R$.
$\frac{v_0^2 \sin^2 \theta}{2g} > \frac{2 v_0^2 \sin \theta \cos \theta}{g}$.
$\frac{\sin^2 \theta}{2} > 2 \sin \theta \cos \theta$.
$\frac{\sin \theta}{2} > 2 \cos \theta$ (assuming $\sin \theta \neq 0$).
$\frac{\sin \theta}{\cos \theta} > 4 \implies \tan \theta > 4$.
$\theta > \arctan(4) \approx 75.96^\circ$. The minimum angle is $76^\circ$.
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