Q. 66

Expert-verifiedFound in: Page 202

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

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 10^{o} 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 10^{o} The skis have a thrust of 200 N.Ski is tilted at 10^{o} after becoming airborne.

We know that work done by a constant force F in moving a body a distance d is given by

W=F d.

Potential energy of a mass at height h is given as mgh.

KE of an object moving with velocity v is given by 1/2mv^{2}

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

we get

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

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