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CIE 9231 2025 November Paper 34 Q2

A Level / CIE / FM

CIE 9231 2025 Nov Paper 34 Paper · Question 2

题目

Problem

A particle PP is projected with speed ums1u\,\mathrm{m\,s^{-1}} at an angle θ\theta, where tanθ=2\tan\theta=2, above the horizontal from a point OO on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of PP from OO at a time tst\,\mathrm{s} are denoted by xmx\,\mathrm{m} and ymy\,\mathrm{m} respectively.

(a) Use the equation of the trajectory given in the list of formulae (MF 19) to show that

y=2x25x2u2.y=2x-\frac{25x^2}{u^2}.
(1)

(b) In the subsequent motion, PP passes through the point with coordinates (8,12)(8,12). The particle then hits a fixed vertical barrier 7m7\,\mathrm{m} high that is at a horizontal distance of DmD\,\mathrm{m} from the point of projection.

Find the set of possible values of DD.

(5)
题目中文翻译

质点 PP 从水平面上的点 OO 以速度 ums1u\,\mathrm{m\,s^{-1}}、与水平方向成角 θ\theta 的方向斜向上抛出,其中 tanθ=2\tan\theta=2,随后仅在重力作用下运动。在 tst\,\mathrm{s} 时,PP 相对于 OO 的水平位移和竖直位移分别记为 xmx\,\mathrm{m}ymy\,\mathrm{m}

(a) 使用公式表(MF 19)中给出的轨迹方程,证明

y=2x25x2u2.y=2x-\frac{25x^2}{u^2}.

(b) 在随后的运动中,PP 经过坐标为 (8,12)(8,12) 的点。该质点随后撞上一面高为 7m7\,\mathrm{m} 的固定竖直挡板;挡板距投射点的水平距离为 DmD\,\mathrm{m}

DD 的可能取值集合。

解答