With what velocity a ball be projected vertically so that the distance covered by it in 5$^{th}$ second is twice the distance it covers in its 6$^{th}$ second ($g = 10\ m/s^{2}$)

Question

Dr. Abubacker · Accepted Answer

(c) $h_{n^{th}} = u - \frac{g}{2}(2n - 1)$
$h_{5^{th}} = u - \frac{10}{2}(2 	imes 5 - 1) = u - 45$
$h_{6^{th}} = u - \frac{10}{2}(2 	imes 6 - 1) = u - 55$
Given $h_{5^{th}} = 2 	imes h_{6^{th}}.$ By solving we get $u = 65\ m/s$

Q. With what velocity a ball be projected vertically so that the distance covered by it in 5$^{th}$ second is twice the distance it covers in its 6$^{th}$ second ($g = 10\ m/s^{2}$)

Explanation

Reviewed by Dr. Abubacker

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