Question on: WAEC Physics - 2011
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]
A
0.75m
B
2.25m
C
8.22m
D
15.00m
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Correct Option: C
S = U\(\sqrt{\frac{2h}{g}}\)
= 15\(\sqrt{\frac{2 \times 1.5}{10}}\)
= 8.22m
= 15\(\sqrt{\frac{2 \times 1.5}{10}}\)
= 8.22m
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