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 â€˜Mythbusters waterslideâ€™ 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. Adamâ€™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

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

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