This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: salazar (ggs332) – HW02 – TSOI – (58160) 1 This printout should have 25 questions. Multiplechoice 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 + v 2 y , where v 2 y = v 2 y 2 gh · parenleftbigg h 4 parenrightbigg = (1 1 4 ) v 2 y . 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 2 gh 0 = v 2 y 2 gh ⇒ v y = 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 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 max height = 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 , we can find the initial projection angle. θ = tan − 1 v y 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 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 b w 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....
View
Full
Document
This note was uploaded on 11/03/2010 for the course PHYSICS 58160 taught by Professor Tsoi during the Spring '10 term at University of Texas.
 Spring '10
 TSOI
 Physics

Click to edit the document details