Question

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

• A. $58.8ms^{- 1}$
• B. $49ms^{- 1}$
• C. $65ms^{- 1}$
• D. $19.6ms^{- 1}$

Answer 1

Answer: (C)

$h_{n^{t h}}=u-\frac{g}{2}\left(2 n - 1\right)$
$h_{5^{t h}}=u-\frac{10}{2}\left(2 \times 5 - 1\right)=u-45$
$h_{6^{t h}}=u-\frac{10}{2}\left(2 \times 6 - 1\right)=u-55$
Given $h_{5^{t h}}=2\times h_{6^{t h}}.$ By solving we get $u=65ms^{- 1}$