If we let v_{0} be the velocity of the ball and the projectile, then we'll have:

$\begin{array}{rcl}\mathbf{T}& \mathbf{=}& \mathbf{(}\mathbf{M}\mathbf{+}\mathbf{m}\mathbf{)}\mathbf{g}\mathbf{+}\frac{\mathbf{(}\mathbf{M}\mathbf{+}\mathbf{m}\mathbf{)}{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}}{\mathbf{l}}\\ \frac{\mathbf{(}\mathbf{M}\mathbf{+}\mathbf{m}\mathbf{)}{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}}{\mathbf{l}}& \mathbf{=}& \mathbf{T}\mathbf{-}\mathbf{(}\mathbf{M}\mathbf{+}\mathbf{m}\mathbf{)}\mathbf{g}\\ {\mathbf{v}}_{\mathbf{0}}& \mathbf{=}& \sqrt{\frac{\mathbf{l}\mathbf{[}\mathbf{T}\mathbf{-}\mathbf{(}\mathbf{M}\mathbf{+}\mathbf{m}\mathbf{\left)}\mathbf{g}\mathbf{\right]}}{\mathbf{M}\mathbf{+}\mathbf{m}}}\\ & \mathbf{=}& \sqrt{\frac{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{)}\mathbf{[}\mathbf{600}\mathbf{-}\mathbf{(}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{900}\mathbf{\left)}\mathbf{\right(}\mathbf{9}\mathbf{.}\mathbf{81}\mathbf{\left)}\mathbf{\right]}}{\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{900}}}\end{array}$

A 10.0 kg wood ball hangs from a 1.30 m-long wire. The maximum tension the wire can withstand without breaking is 600 N. A 0.900 kg projectile traveling horizontally hits and embeds itself in the wood ball.

What is the largest speed this projectile can have without causing the cable to break?

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