Answer:
v ≈ 5.602 m/s
Step-by-step 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:
![K = U_(grav)\\(1)/(2) \cdot m \cdot v^(2) = m \cdot g \cdot h](https://img.qammunity.org/2021/formulas/physics/college/l4vcrcr6snzcty2s8soma3jz70bs2fzfbg.png)
Where
and
are the kinetic energy and gravitational potential energy, respectively.
Initial velocity is found by isolating in that equation.
![v = √(2 \cdot g \cdot h)\\v = \sqrt{2\cdot (9.81 (m)/(s^2))\cdot (1.60 m)} \\v \approx 5.602 (m)/(s)](https://img.qammunity.org/2021/formulas/physics/college/mhu0thu3tna0mzr8devuonth093c5h5zs8.png)