# Find initial velocity when a stuntman jump from $1.25 \ m$ height

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.

## 1 answer

The method is correct.

when you write $s=vt$, $s$ is the horizontal distance and $v$ is the horizontal component of the initial velocity (and it happens to be that the initial velocity has only horizontal component but it could have been different), which does not change through the motion, that's why we can write after all, this equation is when the velocity is constant (or when the velocity is the average velocity but this is not the case here). ~ PF

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