Answer:
v ≈ 5.602 m/s
Explanation:
By applying Work-Energy Theorem and Principle of Energy Conservation, the physical model of given system is described below. It is assumed that height at the bottom is zero:
[tex]K = U_{grav}\\\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot h[/tex]
Where [tex]K[/tex] and [tex]U_{grav}[/tex] are the kinetic energy and gravitational potential energy, respectively.
Initial velocity is found by isolating in that equation.
[tex]v = \sqrt{2 \cdot g \cdot h}\\v = \sqrt{2\cdot (9.81 \frac{m}{s^2})\cdot (1.60 m)} \\v \approx 5.602 \frac{m}{s}[/tex]
