HW2 Solution - salazar(ggs332 HW02 TSOI(58160 This...

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salazar (ggs332) – HW02 – TSOI – (58160) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The speed of a projectile when it reaches its maximum height is 1 4 the speed when the projectile is at 1 4 its maximum height. What is the initial projection angle? Hint: First verify the following relation- ships v x = 1 4 radicalBig v 2 x 0 + v 2 y , where v 2 y = v 2 y 0 - 2 gh · parenleftbigg h 4 parenrightbigg = (1 - 1 4 ) v 2 y 0 . Correct answer: 77 . 3956 . Explanation: First find the y component of the initial velocity in terms of the maximum height, h , and gravity, g . v 2 y = v 2 y 0 - 2 gh 0 = v 2 y 0 - 2 gh v y 0 = radicalbig 2 gh The y component of the velocity at 1/ 4 the maximum height is v 2 y (1 / 4 h ) = v 2 y 0 - 1 4 2 gh = 2 gh - 1 4 2 gh v y (1 / 4 h ) = radicalBigg 2 gh parenleftbigg 1 - 1 4 parenrightbigg We are given, speed maxheight = 1 4 speed 1 4 height At the maximum height the speed of the projectile is simply due to the x component of the velocity. Note that the x component of the velocity is constant. Hence v x = 1 4 radicalBig v 2 x + v 2 y (1 / 4 h ) = 1 4 radicalBigg v 2 x + 2 gh parenleftbigg 1 - 1 4 parenrightbigg Solving for v x yields v x = radicalBigg 2 gh ( 1 - 1 4 ) (4) 2 - 1 Finally, since we now have an expression for v x and v y 0 , we can find the initial projection angle. θ = tan 1 v y 0 v x = 2 gh radicalbig (4) 2 - 1 radicalBig 2 gh (1 - 1 4 ) = tan 1 radicalBigg (4) 2 - 1 (1 - 1 4 ) = 77 . 3956 002 (part 1 of 2) 10.0 points A river flows at a speed v r = 4 . 27 km / hr with respect to the shoreline. A boat needs to go perpendicular to the shoreline to reach a pier on the river’s other side. To do so, the boat heads upstream at an angle θ = 34 from the direction to the boat’s pier. Find the ratio of v b to v r , where v r is defined above and v b is the boat’s speed with respect to the water. 1. v b v r = sin 2 θ 2. v b v r = 1 sin θ correct 3. v b v r = sin θ 4. v b v r = cos θ 5. v b v r = 1 tan θ 6. v b v r = 1 cos 2 θ 7. v b v r = 1 cos θ 8. v b v r = 1 sin 2 θ 9. v b v r = cos 2 θ
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salazar (ggs332) – HW02 – TSOI – (58160) 2 10. v b v r = tan θ Explanation: Let : v bs =? , boat relative to shore v ws = v r , water relative to shore v bs = v b , boat relative to water . This is a problem about relative velocity. See the figure below. L v bs v bw v ws θ The velocity of the boat relative to the shore is given by -→ v bs = -→ v bw + -→ v ws . It is easy to see from the figure above that v ws , v bw and v bs form a right triangle if the boat moves northward relative to the earth. Therefore, v ws = v bw sin θ = v bw v ws = 1 sin θ . 003 (part 2 of 2) 10.0 points If the time taken for the boat to cross the river is 10 . 2 min, determine the width of the river. Correct answer: 1 . 07619 km. Explanation: By similar reasoning, we know relative to the shore, the velocity of the boat is v bs = v ws tan θ = 6 . 33054 km / hr .
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