Question

Starting at rest, Tina slides down a frictionless waterslide with a horizontal section at the bottom that is 5.58 ft above the surface of the swimming pool and strikes the water a distance of 15.9 ft away from the end of the slide. Using conservation of energy, what is Tina's initial height on the waterslide? 16.85 x ft above the bottom of the slide Submit Answer Save Progress

Answer 1

$\\ \text{Lets find the Take-off Speed at the end of the slides, using Projectile Motion}\\ \\ \ x(t)=v_0*t\\ \\ \ 15.9 = v_0*t\\ \\ \text{where t is the time of fall.}\\ \\ \text{Now lets see the vertical motion}\\ \\ \ y(t)=5.58-\frac{1}{2}gt^2, y(t)=0, splashdown, t = \sqrt{\frac{11.16}{g}}=1.057\\ \\ \ v_0*\sqrt{\frac{11.16}{g}}=15.9\\ \\ \ v_0=15.0359\\ \\ \text{Because of loss in Potentail Energy there is Motion in the slider, tina, So we can write}\\ \\ \ mgh=\frac{1}{2}mv^2=>9.98*h=\frac{1}{2}*(15.0359)^2=11.3266\\ \\ \text{So net height would be }=11.3266+5.58ft=16.9066ft.$