陳碧利 0 Posted September 5, 2017 Report Share Posted September 5, 2017 在海邊的山崖頂上,以 100 m/s 的水平速度發射一砲彈,當砲彈掉落海面時,其運動方向與海平面夾角45°,已 知重力加速度g =10 m/ s 2,且不計空氣阻力,求山崖頂的高度約為多少 m? 1. 500 2. 750 3. 1000 4. 1250 答案:1 問題1為何水平初速=水平末速??也為何垂直末速亦為100m/s? (跟海平面夾45度有關??) 問題2:水平拋體公式的公式是甚麼? 問題3:下圖中V=Vo+gt(為何Vo=0????) Link to post Share on other sites
陳碧利 0 Posted September 10, 2017 Author Report Share Posted September 10, 2017 有沒高手會的?拜托 Link to post Share on other sites
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