Question

1. 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 1.500 s after leaving the slide. Ignoring friction and air resistance, find the height H in the drawing. The answer is 11.6 m (please explain)

2. An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts down a distance of 12.6 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 4.30 mbelow the edge. How fast is she going just before she lands? The answer is 12 m/s. Please explain!

Answer 1

let the required height will be h1+h2

h1 is the height of the slide

h2 is the height of the end of the slide from the ground

h2 = 0.5*g*t^2 = 0.5*9.81*1.5*1.5 = 11.03 m

velocity at the end of the slide is v= 5/1.5 =3.33 m/s

apply law of conservationof energy

m*g*h1 = 0.5*m*v^2

h1 = 0.5*v^2/g = 0.5*3.33*3.33/9.81 = 0.565 m

then required height is h = 11.03+0.565 = 11.595 m = 11.6 m

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find the velocity at the cliff:
the gravitational force is
Fg = mg*sin25
the friction force is
Ff = μmgcos25
The net force on her is
Fnet = Fg - Ff = mg*sin25 - μmgcos25
Her acceleration is then
a = Fnet/m = ( mg*sin25 - μmgcos25) / m
a = g*sin25 - μgcos25
a = 9.81(0.2413)
a = 2.367 m/s^2

and v = √(2as)
v = √(2*12.6*2.367)
v = 7.723 m/s

the horizontal component of this velocity is
vh = v*cos25 = 7.723*cos25 = 6.99 m/s
the vertical velocity is
vv = v*sin25 = 7.723*sin25 = 3.26 m/s

find the final vertical velocity 4.5 m below the cliff:
vvf^2 = vv^2 + 2gh
vvf^2 = 3.26^2 + 2*9.81*4.3
vvf = 9.74 m/s

and te total final velocity is found by Pythagoras addition of final vertical velocity and horizontal velocity:
vtot = √(6.99^2 + 9.74^2)
vtot = 12.00 m/s <-- ans