A particle is projected horizontally at 15ms\(^{-1}\) from a height of 20m. Calculate the horizontal distance covered by the particle just before hitting the ground. [g= 10ms\(^{-2}\)]
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Correct Answer: Option n
Explanation:
Let R represent the horizontal distance covered at time t.
Using h = ut + \(\frac{1}{2}gt^2\)
but U\(_v\) = 0
h = \(\frac{1}{2} gt^{2}\)
20 = \(\frac{1}{2} \times 10 \times t^2\)
20 = 5t\(^2\)
t\(^2\) = \(\frac{20}{5}\) = 4
t = 2s
Range(R) = U\(_x\) x t
= 15 x 2
= 30m
or
R = u x \(\sqrt{\frac{2h}{g}}\)
= 15 x \(\sqrt{\frac{2 \times 20}{10}}\)
= 15 x \(\sqrt{4}\)
= 15 x 2
= 30m
Let R represent the horizontal distance covered at time t.
Using h = ut + \(\frac{1}{2}gt^2\)
but U\(_v\) = 0
h = \(\frac{1}{2} gt^{2}\)
20 = \(\frac{1}{2} \times 10 \times t^2\)
20 = 5t\(^2\)
t\(^2\) = \(\frac{20}{5}\) = 4
t = 2s
Range(R) = U\(_x\) x t
= 15 x 2
= 30m
or
R = u x \(\sqrt{\frac{2h}{g}}\)
= 15 x \(\sqrt{\frac{2 \times 20}{10}}\)
= 15 x \(\sqrt{4}\)
= 15 x 2
= 30m