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>A stuntman jumped from $1.25 \ \text{m}$ height and, landed at distance $10 \ \text{m}$. Find velocity when he jumped. (Take $\text{g}=10 \ ms^{-2}$)
I had solved it following way.
$$h=\frac{1}{2}gt^2$$
$$=>1.25=5\cdot t^2$$
$$=>t=\frac{1}{2} \ s$$
And, $$s=vt$$
$$v=\frac{s}{t}$$
$$=\frac{10 \ m}{\frac{1}{2} \ s}$$
$$=20 \ ms^{-1}$$
The answer is correct (checked from book answer). But, $v$ is average speed in the following equation.
$$s=vt$$
But, they told me to find initial velocity. That's why I think my answer is correct but, method is wrong so, the whole work is wrong either.
