A particle is projected horizontally at 10 ms\(^{-1}\) from a height of 45m. Calculate the horizontal distance covered by the particle before hitting the ground. [g = 10ms\(^{-1}\)]
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Correct Answer: Option n
Explanation:
\(h = \frac{1}{2} gt^2\) t = \(\sqrt{\frac{2h}{g}}\)
= \(\sqrt{\frac{2 \times 45}{10}}\) = 3s
Horinzntal distance = vt = 10 x 3 = 30m
\(h = \frac{1}{2} gt^2\) t = \(\sqrt{\frac{2h}{g}}\)
= \(\sqrt{\frac{2 \times 45}{10}}\) = 3s
Horinzntal distance = vt = 10 x 3 = 30m