Trong môn trượt tuyết, một vận động viên sau khi trượt trên đoạn đường dốc thì trượt ra khỏi dốc theo phương ngang ở độ cao 100 m so với mặt đất. Người đó bay xa được 180 m trước khi chạm đất. Hỏi tốc độ của vận động viên đó khi rời khỏi dốc là bao nhiêu? Lấy g = 9,8 m/s2.
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Thời gian rơi của vận động viên được tính bằng công thức: $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 100}{9.8}} \approx 4.52 s$
- Vận tốc ban đầu của vận động viên là: $v_0 = \frac{L}{t} = \frac{180}{4.52} \approx 39.82 m/s$
$t = \sqrt{\frac{200}{9.8}} = \sqrt{\frac{1000}{49}} = \frac{10\sqrt{10}}{7}$
$v_0 = \frac{180}{\frac{10\sqrt{10}}{7}} = \frac{180 \cdot 7}{10\sqrt{10}} = \frac{126}{\sqrt{10}} = 12.6\sqrt{10} \approx 39.81 m/s$
Đáp án gần nhất là 40 m/s nhưng nếu làm tròn số, đáp án phải là $v_0=40 m/s$
Nếu ta tính theo đề bài các dữ kiện như sau:
$t = \sqrt{\frac{2h}{g}}=\sqrt{\frac{2*100}{9.8}} = \sqrt{\frac{2000}{98}}=\sqrt{\frac{1000}{49}}=\frac{10\sqrt{10}}{7}$
$v = \frac{x}{t} = \frac{180}{\frac{10\sqrt{10}}{7}} = \frac{180*7}{10\sqrt{10}} = \frac{18*7}{\sqrt{10}} = \frac{126}{\sqrt{10}} = 39.81 m/s$\nGiá trị này rất gần 40 m/s. Tuy nhiên, đề bài có thể có sai sót. Nếu quãng đường bay xa là 190m thì ta có $v = \frac{190}{\frac{10\sqrt{10}}{7}} = \frac{19*7}{\sqrt{10}} = \frac{133}{\sqrt{10}} = 42.06 m/s$\nVậy đáp án chính xác nhất là 42 m/s.
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