USAPhO 1999, Четвъртфинал — Задача 1

• a. 34.1 °

• b. 67.3 m

  The ball lands 67.3 m behind the front end (official: Δx = −67.3 m; either sign okay, 67.4 m also okay).

1. $y_{\max} = 10\ \mathrm{m}$ $\quad v_o = 25\ \mathrm{m/s}$ $\quad a = 2\ \mathrm{m/s^2}$

a. The train's acceleration does not affect the vertical motion.

At the highest point, $v_y = 0$ [2]

$v_y^2 = v_{oy}^2 - 2gy$ [2]

$v_{oy} = \sqrt{2gy_{\max}} = \sqrt{2(9.8\ \mathrm{m/s^2})(10\ \mathrm{m})} = 14\ \mathrm{m/s}$ [2]

Since $v_{oy} = v_o \sin\theta$ [2]

$\theta = \sin^{-1}\left(v_{oy}/v_o\right) = \sin^{-1}\left(14\ \mathrm{m/s} / 25\ \mathrm{m/s}\right) = 34.1^{\circ}$ [2]

b. The time to reach $y_{\max}$ is obtained from $v_y = v_{oy} - gt$ at the highest point

$t = \dfrac{v_{oy}}{g} = \dfrac{14\ \mathrm{m/s}}{9.8\ \mathrm{m/s^2}} = 1.43\ \mathrm{s}$ [5]

The time to land is twice that, or $t = 2.86$ s. [2]

During that time, the ball moves a distance $x_b = v_{oc}t - v_o\cos\theta t$ [2]

where $v_{oc}$ is the velocity of the train with respect to the Earth when the ball is launched.

And the front end of the train moves $x_p = v_{oc}t + \tfrac{1}{2}at^2$ [2]

The difference of those displacements is the ball's displacement with respect to the train.

$\Delta x = x_b - x_p = -v_o\cos\theta t - \tfrac{1}{2}at^2$ [2]

$\Delta x = -25\ \mathrm{m/s}\left(\cos 34.1^{\circ}\right)2.86\ \mathrm{s} - \tfrac{1}{2}\left(2\ \mathrm{m/s^2}\right)\left(2.86\ \mathrm{s}\right)^2 = -67.3\ \mathrm{m}$ [2]

(Either sign okay. 67.4 m also okay. Part b may also be solved in the train's accelerating frame or in an inertial frame with the same velocity with respect to the Earth that the train had when the ball was launched. Any correct method may be used.)