STANDARD Physics Projectile Motion MCQ Question
A ball is thrown from the top of a tower with an initial velocity of 10 m s⁻² at an angle of 30° with the horizontal. If it hits the ground at a distance of 17.3 m from the base of the tower, the height of the tower is (Take g = 10 m s⁻²)
5 m
20 m
15 m
10 m
Correct Answer
Detailed Explanation
For horizontal motion, the range R = u cosθ × t. Solving for t gives t = 2 s. For vertical motion, y = u sinθ × t - 1/2 gt². Substituting the values gives the height y = 10 m.
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