With what velocity a ball be projected vertically so that the distance covered by it in 5 textth second is twice the distance it covers in its 6 textth second (g = 10 m s- 2) .

Question

With what velocity a ball be projected vertically so that the distance covered by it in 5 textth second is twice the distance it covers in its 6 textth second (g = 10 m s- 2) . (A) 58.8ms- 1 (B) 49ms- 1 (C) 65ms- 1 (D) 19.6ms- 1

Tardigrade Admin · Accepted Answer

hnt h=u-(g/2)(2 n - 1) h5t h=u-(10/2)(2 × 5 - 1)=u-45 h6t h=u-(10/2)(2 × 6 - 1)=u-55 Given h5t h=2× h6t h. By solving we get u=65ms- 1

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

$58.8 m s^{- 1}$

$49 m s^{- 1}$

$65 m s^{- 1}$

$19.6 m s^{- 1}$

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

Q. With what velocity a ball be projected vertically so that the distance covered by it in 5th second is twice the distance it covers in its 6th second (g=10ms−2) .

58.8ms−1

49ms−1

65ms−1

19.6ms−1