Question

A 52.4 kg child slides down a water slide with a velocity of 1.1 m/sec at the top. At the bottom of the slide, she is moving horizontally, y=1.5 meters above the water. She splashes into the water d=2 meters to the left of the bottom of the slide.1) Assuming potential energy to be zero at the water level, what is the mechanical energy of the child at the top of the slide? MEo=?. 2) How high is the top of the slide above the bottom of the slide? h =?

Answer 1

Solution-We have

Horizontal distance = v*t = 2m

and Vertical distance = 1.5 = 0.5*9.81*t^2
=>t = sqrt(1.5/(0.5*9.81)) = 0.5530 s

Also
vt = 2.0

=>v=2.0/0.5530 s
=>v = 3.616 m/s Answer
1)Here :

Mechanical energy at the top of the slide = Mechanical energy at the bottom of the slide

= >mgh + 0.5mv^2 = (9.81*1.5*52.4)+(0.5*52.4*(3.616 ^2)) = 1113.64 J
2)Mechanical energy at the top of the slide = Mechanical energy at the bottom of the slide
=>1113.64= 52.4*9.81*h + 0.5*52.4*(1.1^2)
=>h = 2.104 m