$\begin{array}{1 1} zero \\ 34\;m \\ 4\;m \\2\;m \end{array} $

$S_N = u+\large\frac{a}{2} $$(2n-1)$

$S_5= u+ \large\frac{9a}{2}$-------(1)

$S_8= u+ \large\frac{15a}{2}$-------(2)

Solving (1) and (2) we get $a-2 \;ms^{-2}$ and $ u-5\;ms^{-1}$

$S_{15}=u + \large\frac{29a}{2} $

$\quad= 5 +\large\frac{29 \times 2}{2}$$=34\;m$

The value of a and u in above solution won't be negative.

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