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 small point-like object is thrown horieprj 8vbk rp-s:6 vn;ys,l,7mzocq-ekh1:ztc /ntally off of a $50.0\,\mathrm{m}$ high building with an initial speed of $10.0\,\mathrm{m/s}$. At any point along the trajectory there is an acceleration component tangential to the trajectory and an acceleration component perpendicular to the trajectory. How many seconds after the object is thrown is the tangential component of the acceleration of the object equal to twice the perpendicular component of the acceleration of the object? Ignore air resistance.
A. $2.00\,\mathrm{s}$
B. $1.50\,\mathrm{s}$
C. $1.00\,\mathrm{s}$
D. $0.50\,\mathrm{s}$
E. The building is not high enough for this to occur.
参考答案: A
本题详细解析:
The only acceleration is gh88ej+zo0,kn(xg7k3k* i w iebfwag 7ravity, $\vec{a} = \vec{g}$, which points vertically downward. Let's use $\vec{g} = (0, -g)$.
The velocity vector at time $t$ is $\vec{v}(t) = (v_x, v_y) = (10.0, -gt)$.
The tangential acceleration $a_t$ is the projection of $\vec{a}$ onto $\vec{v}$: $a_t = (\vec{a} \cdot \vec{v}) / |\vec{v}|$.
$\vec{a} \cdot \vec{v} = (0, -g) \cdot (10, -gt) = 0 + g^2t = g^2t$.
The perpendicular (normal) acceleration $a_n$ is the component of $\vec{a}$ perpendicular to $\vec{v}$.
We can use the Pythagorean theorem for acceleration components: $a^2 = a_t^2 + a_n^2$.
Since $a=g$, we have $g^2 = a_t^2 + a_n^2$.
We are given the condition $a_t = 2a_n$.
Substitute $a_n = a_t / 2$ into the Pythagorean relation:
$g^2 = a_t^2 + (a_t / 2)^2 = a_t^2 + a_t^2 / 4 = \frac{5}{4}a_t^2$
$a_t^2 = \frac{4}{5}g^2 \implies a_t = \frac{2}{\sqrt{5}}g$.
Now use the expression for $a_t$: $a_t = (\vec{a} \cdot \vec{v}) / |\vec{v}| = g^2t / \sqrt{10^2 + (gt)^2}$.
$\frac{2}{\sqrt{5}}g = g^2t / \sqrt{100 + g^2t^2}$.
Square both sides: $\frac{4}{5}g^2 = g^4t^2 / (100 + g^2t^2)$.
$\frac{4}{5} = g^2t^2 / (100 + g^2t^2)$
$4(100 + g^2t^2) = 5g^2t^2$
$400 + 4g^2t^2 = 5g^2t^2$
$400 = g^2t^2 \implies t = \sqrt{400 / g^2} = 20 / g$.
Using $g \approx 10\,\mathrm{m/s^2}$, $t = 20 / 10 = 2.00\,\mathrm{s}$.
(We must check if it hits the ground: $t_{fall} = \sqrt{2y/g} = \sqrt{2(50)/10} = \sqrt{10} \approx 3.16\,\mathrm{s}$. Since $2.00\,\mathrm{s} < 3.16\,\mathrm{s}$, this event occurs.)
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