2. a boy and girl are climbing a tree. the girl is $10\text{ m}$ above the ground and throws a ball horizontally at $18\text{ ms}^{-1}$. the boy is $12\text{ m}$ above the ground and can only throw her ball horizontally at $15\text{ ms}^{-1}$. whose ball lands furthest from the tree?

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{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Projectile Motion", "Horizontal Launch" ], "new_concepts": [], "current_concepts": [ "Projectile Motion", "Horizontal Launch", "Kinematics" ] } ## Calculate the horizontal range of the girl's ball $$ \begin{aligned} &h_g = 10\text{ m},\quad v_g = 18\text{ ms}^{-1},\quad g = 9.8\text{ ms}^{-2}\ &t_g = \sqrt{\frac{2h_g}{g}} = \sqrt{\frac{2 \times 10}{9.8}} \approx 1.4286\text{ s}\ &d_g = v_g \times t_g = 18 \times 1.4286 \approx 25.71\text{ m} \end{aligned} $$ ## Calculate the horizontal range of the boy's ball $$ \begin{aligned} &h_b = 12\text{ m},\quad v_b = 15\text{ ms}^{-1}\ &t_b = \sqrt{\frac{2h_b}{g}} = \sqrt{\frac{2 \times 12}{9.8}} \approx 1.5649\text{ s}\ &d_b = v_b \times t_b = 15 \times 1.5649 \approx 23.47\text{ m} \end{aligned} $$ ## Compare the horizontal distances $$ \begin{aligned} &d_g \approx 25.71\text{ m} > d_b \approx 23.47\text{ m} \end{aligned} $$ The girl's ball lands furthest from the tree. { "subject": "Natural Science", "question_type": "Word Problem", "knowledge_point": [ "Natural Science", "Physics", "Projectile Motion" ] }

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Explanation:

Calculate the horizontal range of the girl's ball

Calculate the horizontal range of the boy's ball

Compare the horizontal distances

Answer: