Answer to Question #182043 in Mechanics | Relativity for buddy

Question

A child on a slide is traveling 1.02 m/s 2.10 m off the ground. How fast is the child going at the bottom?

Answer 1

From the law of energy conservation,

$\displaystyle mgh + \frac{mv_0^2}{2} = \frac{mv^2}{2}$

$2gh + v_0^2 = v^2$

$v = \sqrt{2gh+v_0^2} = \sqrt{2 \cdot 9.81 \cdot2.1+1.02^2}=6.5\; m/s.$

Answer: 6.5 m/s