Question

20．從地面豎直上拋的小球，空氣阻力不計(jì)，在拋出后的時(shí)刻t₁和時(shí)刻t₂的位移相同，則它拋出時(shí)的初速度大小為$\frac{（{t}_{1}^{\;}+{t}_{2}^{\;}）g}{2}$，在時(shí)刻t₁時(shí)離地面的高度為$\frac{{t}_{1}^{\;}{t}_{2}^{\;}g}{2}$．

Answer 1

分析 根據(jù)豎直上拋運(yùn)動(dòng)的對(duì)稱(chēng)性，${t}_{1}^{\;}$時(shí)刻到最高點(diǎn)的時(shí)間$\frac{{t}_{2}^{\;}-{t}_{1}^{\;}}{2}$，從而即可得出從拋出點(diǎn)到最高點(diǎn)的時(shí)間，根據(jù)速度時(shí)間關(guān)系得到初速度，再根據(jù)位移時(shí)間關(guān)系得出${t}_{1}^{\;}$時(shí)刻距地面的高度．

解答 解：根據(jù)豎直上拋運(yùn)動(dòng)的對(duì)稱(chēng)性，${t}_{1}^{\;}$時(shí)刻到最高點(diǎn)的時(shí)間$\frac{{t}_{2}^{\;}-{t}_{1}^{\;}}{2}$
從拋出點(diǎn)到最高點(diǎn)的時(shí)間$t={t}_{1}^{\;}+\frac{{t}_{2}^{\;}-{t}_{1}^{\;}}{2}=\frac{{t}_{1}^{\;}+{t}_{2}^{\;}}{2}$
拋出時(shí)的初速度$v=gt=\frac{（{t}_{1}^{\;}+{t}_{2}^{\;}）}{2}g$
${t}_{1}^{\;}$時(shí)刻離地面的高度$h=v{t}_{1}^{\;}-\frac{1}{2}g{t}_{1}^{2}=\frac{（{t}_{1}^{\;}+{t}_{2}^{\;}）g}{2}{t}_{1}^{\;}-\frac{1}{2}g{t}_{1}^{2}$=$\frac{{t}_{1}^{\;}{t}_{2}^{\;}g}{2}$
故答案為：$\frac{（{t}_{1}^{\;}+{t}_{2}^{\;}）g}{2}$           $\frac{{t}_{1}^{\;}{t}_{2}^{\;}g}{2}$

點(diǎn)評(píng) 解決本題的關(guān)鍵是掌握豎直上拋運(yùn)動(dòng)的規(guī)律，注意豎直上拋運(yùn)動(dòng)的加速度等于重力加速度，是勻變速直線運(yùn)動(dòng)，速率和時(shí)間具有對(duì)稱(chēng)性．