In the famous TV program, Mythbusters, Adamand Jamie tested a waterslide. Please
ID: 1858526 • Letter: I
Question
In the famous TV program, Mythbusters, Adamand Jamie tested a waterslide. Please referto
below ‘Mythbusters waterslide’ in YouTube.
Part 1: http://www.youtube.com/watch?v=AMkW_vg6ygA
Part 2: http://www.youtube.com/watch?v=1T8xKGcmUuY
Part 3: http://www.youtube.com/watch?v=TGAnWgvl7SE
Part 4: http://www.youtube.com/watch?v=0OTrb_RwHz8
The geometry ofthe waterslide used in the TV show is(1m=3.28ft):
Slide length= 61m; Slope Angle=24 °; Ramp height=3.66m
The lubricantis used properly on the surface ofthe slide,therefore the friction isignored. Adam’s
weightis 70kg and he istreated as a particle. Adamstartssliding fromthe top.
Case 1: If Ramp angel=30 °
Using theoretical methods to determine: i) themagnitude and direction ofthe velocity
of Adamwhen he justflies away fromthe ramp; ii)the horizontal distance that Adam
flies.
Case 2:to change the ramp angle when keep other parameters unchanged:
Using theoretical methodsto determine themagnitude ofthe ramp angle which will
resultin the max flying distance.
SORRY FOR THE PIC DIDN'T WANT TO SAVE THE ORIGINAL
Explanation / Answer
1)according to conservation of energy
we know
1/2 mv^2 = mg (h1-h2)
=>v=sqrt(2g(h1-h2))
let the extra edge unknown be a
=>a/sin(60)=3.66/sin(54)
=>a =3.91
total length of the slant part =3.91+61 =64.91m
=>h1-h2=64.91 * sin (24) -3.66 =22.74 m
=>v =sqrt(2*9.8(22.74)) =21.11 m/s 30 deg from ground (in the direction of wedge)
horizontal range = sqrt(2*3.66/g)*21.11 *cos(30)+ v^2 *sin(2*30)/g = 3.69 +45.69 =49.3827 m away from the wedge
since, horizontal range is distance travelled while projectile and during free fall
2)do similarly when the angle is changed