# ps03_solns - Physics 1311 Spring 2010 PS #3 Solutions Page...

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Physics 1311 Spring 2010 PS #3 Solutions Page 1 of 9 Physics 1311 Problem Set #3 Solutions Problem 1: Stones are thrown horizontally, with the same initial velocity, from the tops of two diFerent buildings, A and B. The stone from building A lands 2 times as far from the base of the building as does the stone from building B. What is the ratio of building A’s height to building B’s height? This problem is an exercise in proportional reasoning—using the equations we’ve learned to solve problems without numbers. The stones are thrown horizontally, which means that their initial vertical velocity is zero. Since there’s no force in the horizontal direction, the stones move horizontally with the same constant speed v . This gives us a relationship between the horizontal range of the stones and the time, x = vt . In the vertical direction, the stones experience the vertical force of gravity. Since the initial vertical velocity is zero, we get Δ p y = - mg Δ t = v y = - gt. Then, using the position update formula in the vertical direction, Δ y = v y,avg Δ t = Δ y = v y 2 Δ t = - 1 2 gt 2 . Thus the height of the building and the time it takes the stone to fall are related by h = 1 2 2 . We combine this with the previous relation, to obtain h = 1 2 g ± x v ² 2 = h = g 2 v 2 x 2 . This relationship says that the height of the building is proportional to the square of the stone’s range. (The other factors in this expression are constants; they don’t change from building A to building B.) So if the stone from building A lands 2 times as far away, this must mean that building A is 4 times as tall as building B.

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Physics 1311 Spring 2010 PS #3 Solutions Page 2 of 9 Problem 2: A projectile is launched with an initial speed of v 0 at an angle θ above the horizontal. It lands at the same level from which it was launched. What is the magnitude of its average velocity between launch and landing? θ θ Between launch and landing, there is no net vertical displace- ment, so the average of the vertical component of the velocity is zero. The horizontal velocity is a constant v 0 cos θ through- out the motion, so this is the projectile’s average velocity be- tween launch and landing. Another way to approach this question is to Fnd the aver- age velocity vector by averaging the initial and Fnal velocity vectors. The initial velocity vector, in component form, is ± v i =( v 0 cos θ ) ˆx +( v 0 sin θ ) ˆy . The Fnal velocity vector will have a magnitude v 0 (like the initial speed) but will be directed at an angle θ below the horizontal: ± v f v 0 cos θ ) - ( v 0 sin θ ) . When we average these two vectors, the vertical components cancel out, and the (equal) horizontal components average together to produce v 0 cos θ . Problem 3: Olympic diver Matthew Mitcham springs upward from a diving board that is 3.30 m above the water. He enters the water at a 76.0 degree angle with respect to the water surface, at a speed of 9.20 m/s. (a) Determine the magnitude of his initial velocity. (b) Determine the direction of his initial velocity, in terms of degrees relative to horizontal. (c) Determine his maximum height above the water.
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## This note was uploaded on 02/21/2010 for the course PHYS 1311 taught by Professor Wiegert during the Spring '10 term at University of Georgia Athens.

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ps03_solns - Physics 1311 Spring 2010 PS #3 Solutions Page...

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