## Fundamentals of Physics Extended (10th Edition)

$d=48.5m$
To find the velocity of the ball after the fall, use an equation relating distance, initial velocity, acceleration, and final velocity, which is $$v_f^2=v_o^2+2a\Delta y$$ Solve for $v_f$ to get $$v_f=\sqrt{v_o^2+2a\Delta y}$$ Substitute known values of $v_o=0.00m/s$, $\Delta y=-5.20m$, and $a=-9.81m/s^2$ yields a final velocity of $$v_f=\sqrt{2(-9.80m/s^2)(-5.20m)}=10.1m/s$$Since the ball is traveling at a constant velocity, the equation for velocity can be used. $$v=\frac{\Delta x}{\Delta t}$$ Solve for $\Delta x$ to get $$\Delta x=v\Delta t$$ Substitute known values of $v=10.1m/s$ and $t=4.80s$ to get a depth of $$x=(10.1m/s)(4.80s)=48.5m$$