本题目来源于试卷: 2007美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A ball of mass $m$ is launched into the air. Ignore air resistance, but assume that there is a wind that exerts a constant force $F_{o}$ in the $-x$ direction. In terms of $F_{0}$ and the acceleration due to gravity $g$, at what angle above the positive x-axis must the ball be launched in order to come back to the point from which it was launched?
A. $\tan^{-1}(F_0/mg)$
B. $\tan^{-1}(mg/F_0)$
C. $\sin^{-1}(F_0/mg)$
D. the angle depends on the launch speed
E. no such angle is possible
参考答案: B
本题详细解析:
This is a projectile motion problem with a constant acckz1hagb0qx6 at)xs3/g7/qj x eleration vector.
The acceleration componjtxbqq)k /gaax7s/x16g0z3hents are:
$a_x = -F_0 / m$
$a_y = -g$
The total acceleration vector is $\vec{a} = (-F_0/m)\hat{i} - g\hat{j}$.
The displacement after time $t$ is $\Delta\vec{r} = \vec{v}_0 t + \frac{1}{2}\vec{a}t^2$.
For the ball to return to the launch point, the total displacement $\Delta\vec{r}$ must be zero at some time $T > 0$.
$0 = \vec{v}_0 T + \frac{1}{2}\vec{a}T^2$.
Since $T > 0$, we can divide by $T$: $0 = \vec{v}_0 + \frac{1}{2}\vec{a}T \implies \vec{v}_0 = -\frac{1}{2}\vec{a}T$.
This equation tells us that the initial velocity vector $\vec{v}_0$ must be in the *exact opposite direction* of the acceleration vector $\vec{a}$.
The acceleration vector $\vec{a}$ points into the 3rd quadrant (negative $x$, negative $y$).
The launch velocity $\vec{v}_0$ must therefore point into the 1st quadrant (positive $x$, positive $y$).
The angle $\theta$ of the launch $\vec{v}_0$ (above the +x axis) is determined by the components of $-\vec{a}$:
$\tan(\theta) = (v_0)_y / (v_0)_x = (-a_y) / (-a_x) = (-(-g)) / (-(-F_0/m)) = g / (F_0/m) = mg/F_0$.
Therefore, $\theta = \tan^{-1}(mg/F_0)$.
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