A particle is projected horizontally at 10 ms\(^{-2}\) from the top of a tower 20 M high. Calculate the horizontal distance travelled by the particle when it hits the level ground. [g= 10 ms\(^{-2}\)
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Correct Answer: Option n
Explanation:
R = U\(\sqrt{\frac{2H}{g}} = 10\sqrt{\frac{2 \times 20}{10}}\) = 20
R = U\(\sqrt{\frac{2H}{g}} = 10\sqrt{\frac{2 \times 20}{10}}\) = 20