Answer
$216m/s$.
Work Step by Step
All terms must be in SI units. First, convert the time to seconds. This time is equal to $\Delta t = 6.50ms=0.00650s$. Next, convert the acceleration to $m/s^2$. Do this by multiplying the coefficient by the acceleration of gravity to get $a=-3400(9.80m/s^2)=-33300m/s^2$ To find the change in velocity, use the equation relating change in velocity, acceleration, and change in time, which is $$a=\frac{\Delta v}{\Delta t}$$ Solve for $\Delta v$ to get $$\Delta v=a\Delta t$$ Substitute known values of $a=-33300m/s^2$ and $\Delta t =0.00650s$ to get a change in velocity of $$\Delta v=(-33300m/s^2)(0.00650s)=-216m/s$$ Therefore, the initial velocity of the recorder must be $216m/s$.
