An athlete jumps a horizontal distance of $7.18$m. He was airborne for $2.29$ seconds. The acceleration due to gravity is taken as $9.81$ms$^{-2}$. Assume that air resistance is negligible. Calculate the take-off speed in ms$^{-1}$
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$\begingroup$The horizontal component $v_1$ of the velocity is an unchanging $\dfrac{7.18}{2.29}$ metres per second.
Maximum height is reached after $\dfrac{2.29}{2}$ seconds. If $v_2$ is the initial vertical component of the velocity, then the vertical component of the velocity, after $t$ seconds, is $v_2-9.81t$ (for $t\le 2.29$). This vertical component of the velocity reaches $0$ at time $\frac{2.29}{2}$. It follows that $$v_2-9.81\frac{2.29}{2}=0.$$ Now we know $v_1$ and $v_2$. The initial speed is $\sqrt{v_1^2+v_2^2}$.
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