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Question #9ed21

1 Answer

Jan 25, 2018

$65 \frac{m}{s}$ $65 m/s$

Explanation:

Suppose,the ball was projected vertically with velocity $u$ $u$

So,distance( $s$ $s$ ) covered in $5 t h$ $5th$ second is $u - \frac{1}{2} g (2 \cdot 5 - 1)$ $u- 1/2g(2*5-1)$

and, in $6 t h$ $6th$ second is $u - \frac{1}{2} g (2 \cdot 6 - 1) = (h)$ $u-1/2g(2*6-1) = (h)$

Now,given,, $s = 2 h$ $s=2h$

So, equating the obtained equations,we get, $u = \frac{13 g}{2}$ $u=(13g)/2$

Now,if $g = 10 \frac{m}{s^{2}}$ $g= 10 m/s^2$ ,then $u = 65 \frac{m}{s}$ $u=65m/s$

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