Question: With what velocity a ball be projected vertically so that the distance covered by it in 5<sup>th</su...

Question

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

Answer 1

Solution

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

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

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

Given $h_{5^{th}} = 2 \times h_{6^{th}}$ .By solving (i) & (ii) we get $u = 65\mspace{6mu} m/s$