With what velocity a ball be projected vertically so that distance covered by it in 5^{th} second is

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2.Motion in Straight Line

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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.........$m/s$ $(g = 10\,m/{s^2})$

A

$58.8$

B

$49$

C

$65$

D

$19.6$

Solution

(c) ${h_{{n^{th}}}} = u – \frac{g}{2}(2n – 1)$

${h_{{5^{th}}}} = u – \frac{{10}}{2}(2 \times 5 – 1) = u – 45$

${h_{{6^{th}}}} = u – \frac{{10}}{2}(2 \times 6 – 1) = u – 55$

Given ${h_{{5^{th}}}} = 2 \times {h_{{6^{th}}}}$.

By solving we get $u = 65\;m/s$

Standard 11

Physics

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