【題目】在不同高度同時(shí)釋放兩個(gè)鉛球不計(jì)空氣阻力,則在均未落地前,兩者

在任一時(shí)刻具有相同的加速度、位移和速度

落地的時(shí)間間隔取決于兩鉛球釋放時(shí)的高度

在第1 s內(nèi)、第2 s內(nèi)、第3 s內(nèi)位移之比都為149

兩鉛球的距離和速度差都越來(lái)越大

A只有①②正確 B只有①②③正確

C只有①③④正確 D①②③④都正確

【答案】A

【解析】?jī)蓚(gè)鉛球都做自由落體運(yùn)動(dòng),所以加速度都等于重力加速度,即加速度相等;由于同時(shí)釋放,所以運(yùn)動(dòng)時(shí)間相同;根據(jù)公式h=gt2可得發(fā)生的位移相等,任意時(shí)刻的速度v=gt相等,正確;兩鉛球落地的時(shí)間間隔為Δt=,即取決于兩鉛球下落的高度,正確;在第1 s內(nèi)、第2 s內(nèi)、第3 s內(nèi)位移之比都為135,錯(cuò)誤;因?yàn)槭峭瑫r(shí)釋放的,所以兩鉛球下落過(guò)程中相同時(shí)間內(nèi)的位移和速度都相等,故距離差是一個(gè)定值,即釋放前的高度差,錯(cuò)誤。顯然只有選項(xiàng)A正確。

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