A child slides down a slippery slide (neglect friction)of length 3.0 m and angle of inclination 32 $\circ$ above the horizontal.How fast is she going when she leaves the end of the slide?

Question

Quizplus · Accepted Answer

Using conservation of energy, the potential energy at the top of the slide is converted to kinetic energy at the bottom:
mgh = 1/2 mv^2
where m is the mass of the child, g is the gravitational acceleration (9.8 m/s^2), h is the height of the slide (h = 3*sin(32°) = 1.59 m), and v is the velocity of the child at the bottom of the slide.
Solving for v:
v = sqrt(2gh) = sqrt(2*9.8*1.59) = 5.6 m/s
Therefore, the answer is choice C.

A Child Slides Down a Slippery Slide (Neglect Friction)of Length ∘\circ∘

A Child Slides Down a Slippery Slide (Neglect Friction)of Length $\circ$