本题目来源于试卷: Space Time & Motion A.1 Kinematics，类别为 IB物理学
[单选题]
A coin is tossed vertically upwards from a height of 1 2 gmf:lt182d:jqzh nx.5 bik/4 -dx e2c,o(78 k+ag/oxemn;e 7ypxfgbx m. What is the maximum height reached above the ground if the coin was projected with an initial velocix27 x+b-xfid; k e,gemxypn 87(eg/a co/4o bkty of $3.0\,m\,s^{-1}$? (Take $g=10\,m\,s^{-2}$)
A. 0.45 m
B. 1.50 m
C. 1.95 m
D. 6.00 m
参考答案: C
本题详细解析:
Maximum height occurs when the final speed $v$ is zero. The height reached above hand $s = \frac{v^2 - u^2}{2a} = \frac{0 - 9}{-20} = 0.45 \text{ m} $ To find maximum height above the ground, 0.45 m + 1.5 m = 1.95 m
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