A coin is pushed from the edge of a laboratory bench with a horizontal velocity of 15.0 ms-1. If the height of the bench from the floor is 1.5m, calculate the distance from the foot of the bench to the point of impact with the floor [g = 10ms-2]
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Correct Answer: Option C
Explanation:
S = U\(\sqrt{\frac{2h}{g}}\)
= 15\(\sqrt{\frac{2 \times 1.5}{10}}\)
= 8.22m
S = U\(\sqrt{\frac{2h}{g}}\)
= 15\(\sqrt{\frac{2 \times 1.5}{10}}\)
= 8.22m