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.........$m/s$ $(g = 10\,m/{s^2})$
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.........$m/s$ $(g = 10\,m/{s^2})$
$58.8$
$49$
$65$
$19.6$
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.........$m/s$ $(g = 10\,m/{s^2})$
${h_{{n^{th}}}} = u - \frac{g}{2}(2n - 1)$
${h_{{5^{th}}}} = u - \frac{{10}}{2}(2 \times 5 - 1) = u - 45$
${h_{{6^{th}}}} = u - \frac{{10}}{2}(2 \times 6 - 1) = u - 55$
Given ${h_{{5^{th}}}} = 2 \times {h_{{6^{th}}}}$.
By solving we get $u = 65\;m/s$
Other Language