Physics help! ?
A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person hits the water 5.00 m from the end of the slide in a time of 0.584 s after leaving the slide. Ignore friction and air resistance, find the height H in the drawing.
There is no drawing within this problem it is a typo, sorry about the confusion!
- PhilomelLv 74 weeks ago
h=1/2 gt^t h=.5*9.8*.584^2=1.67m
- NCSLv 74 weeks ago
If H is measured from the bottom of the slide to the water,
H = ½gt² yields
H = ½ * 9.81m/s² * (0.584s)² = 1.67 m
If H is measured from the top of the slide, then
H = h₀ + v²/(2g) = 1.67m + (5.00m/0.584s)²/19.6m/s² = 5.41 m
- oubaasLv 74 weeks ago
I answered this question yet, but since i didn't use a magnifying glass i misread the pic reading 5 instead of H :
This is the reason why i just computed the initial speed Vo instead of the requested H
falling height h' = g/2*t^2 = 4.903*0.584^2 = 1.672 m
horizontal velocity V = d/t = 5.00/0.584 = 8.562 m/sec
m/2*V^2+m*g*h' = m*g*h
mass m calcels
h = (V^2/2+g*h') / g = (8.562^2/2+9.806*1.672)/9.806 = 5.41 m
- JimLv 71 month ago
y(t) = ½gt² + v₀t + y₀, v=at and d=vt are the basic formula you need to know.
Standard gravity 'g' is -9.80665 m/s²
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- Andrew SmithLv 71 month ago
Your drawing is done in white ink on white paper.
a) find the horizontal speed ( v = 5/0.584)
b) find the height that gives that speed v^2 = 2gh -> h = v^2 / ( 2*g)