qz4sol_3705sp10

qz4sol_3705sp10 - equation at the starting point and at the...

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TA: Tomoyuki Nakayama Monday, February 15 th , 2010 PHY 2048: Physic 1, Discussion Section 3705 Quiz 4 (Homework Set #6) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ A 60 kg skier starts from rest at height H = 30 m above the end of a ski-jump ramp and the leaves the ramp at angle θ = 25.0º. Neglect the effects of air resistance and assume the ramp is frictionless. a) What is the speed of the skier when he/she leaves the ramp? Since the normal force on the skier is always perpendicular to the skier’s motion, only the gravitational force does work on the skier. Thus the mechanical energy of the skier is conserved. Applying the energy conservation
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Unformatted text preview: equation at the starting point and at the end of the ramp, we get E i = E f mgH = (1/2)mv 2 v = (2gH) = 24.2 m/s b) What is the maximum height h of his/her jump above the end of the ramp? The vertical component of the launching velocity is v 0y = vsin = 10.2 m/s At the peak of the jump, the vertical component of the velocity is zero. The kinematic equation which relates the velocity of a projectile to its displacement yields v y 2 - v 0y 2 = -2gy 2 v 0y 2 = -2gh h = v 0y 2 / 2g = 5.33 m c) If s/he increased his/her weight by putting on a backpack, would h then be greater, less or the same? The launching velocity does not depend on the weight of the skier, and the projectile motion does not depend on the mass of a projectile. Therefore, h is the same....
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