Sam (75 kg) takes off up a 50-m-high, 10 frictionless slope on his jet-powered skis. The skis have a thrust of 200 N. He keeps his skis tilted at 10o after becoming airborne, as shown in FIGURE CP8.66. How far does Sam land from the base of the cliff?
Sam will land at a distance from cliff is : 57.41 meters.
Mass = 75 kg
Takes off at 50 m heightAngle of projection 10o The skis have a thrust of 200 N.Ski is tilted at 10o after becoming airborne.
We know that work done by a constant force F in moving a body a distance d is given by
Potential energy of a mass at height h is given as mgh.
KE of an object moving with velocity v is given by 1/2mv2
The displacement S of an object moving with acceleration and initial speed u is given by
From the law of conservation of energy
Kinetic energy at the release point is equal to work done minus potential energy
Substitute the given values and calculate v
Now find time it takes for Sam to touch down
Solve the quadratic equation
t= - 2.912 s, and 3.501 s
Drop the negative value so t=3.501 sec.
So total distance will be horizontal component of velocity multiplied by time
In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in FIGURE P8.51.
a. Suppose the ring rotates once every 4.5 s. If a rider’s mass is 55 kg, with how much force does the ring push on her at the top of the ride? At the bottom? b. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?
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