A 31.7 kg kid, initially at rest, slides

down a frictionless water slide at 53.2º.

How fast is she moving 3.45 s later?

34.55m/s

Explanation:

Now from Newton's law of motion;

V = U + a*t

V - final velocity

U - initial velocity = 0m/s

a - acceleration

t - time

Note that a = g/cos53.2°

Now g is acceleration of free fall due to gravity

This is a vertical acceleration since the course of motion is along a plane inclined at 53.2° to the height of the pool.

Also from trigonometry identity

cos 53.2° = vertical component a / acceleration (cos a = adjacent/hypothenus)

If we assume g = 9.8m/S2 { a generally given value}

Substituting t = 3.45s as the time of consideration, we have;

V = 0 + (9.8/cos53.2°) * 3.45

= - 34.55m/s

Note the - ve sign is just telling us this velocity is causing the object to move down.

Hence V = 34.55m/s