md1 - Version 072/ABACA – midterm 01 – Turner –...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 072/ABACA – midterm 01 – Turner – (58120) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The speed of an arrow fired from a compound bow is about 64 m / s. An archer sits astride his horse and launches an arrow into the air, elevating the bow at an angle of 53 ◦ above the horizontal and 1 . 4 m above the ground. The acceleration of gravity is 9 . 81 m / s 2 . b b b b b b b b b b b b b b b b b b b b 6 4 m / s 5 3 ◦ 1 . 4 m range What is the arrow’s range? Assume: The ground is level. Ignore air resistance. 1. 261.907 2. 24.4111 3. 323.976 4. 389.927 5. 53.9349 6. 537.785 7. 402.411 8. 160.421 9. 490.125 10. 484.748 Correct answer: 402 . 411 m. Explanation: Let : v o = 64 m / s , θ = 53 ◦ , h = 1 . 4 m , and g = 9 . 81 m / s 2 . b b b b b b b b b b b b b b b b b b b b v o θ h range In the projectile motion, we have x = ( v o cos θ ) t t = x v o cos θ horizontally, and y = h + ( v o sin θ ) t − 1 2 g t 2 vertically. Thus y = h + x v o sin θ v o cos θ − 1 2 g x 2 v 2 o cos 2 θ = 0 when the arrow lands. Thus 2 v 2 o h cos 2 θ + 2 x v 2 o sin θ cos θ − g x 2 = 0 − 2 v 2 o h cos 2 θ − x v 2 o sin2 θ + g x 2 = 0 x = v 2 o sin 2 θ ± radicalBig v 4 o sin 2 2 θ + 8 g v 2 o h cos 2 θ 2 g . Since v 4 o sin 2 2 θ + 8 g v 2 o h cos 2 θ = (64 m / s) 4 (sin 2 106 ◦ ) + (8) (9 . 81 m / s 2 ) × (64 m / s) 2 (1 . 4 m) (cos 2 53 ◦ ) = 1 . 56655 × 10 7 m 2 / s 2 , then x = (64 m / s) 2 (sin106 ◦ ) 2 (9 . 81 m / s 2 ) + radicalbig 1 . 56655 × 10 7 m 2 / s 2 2 (9 . 81 m / s 2 ) = 402 . 411 m . 002 10.0 points Consider the plot describing motion along a straight line with an initial position of 10 m. Version 072/ABACA – midterm 01 – Turner – (58120) 2 − 2 − 1 1 2 3 4 1 2 3 4 5 6 7 8 9 b b b b b time (s) velocity(m/s) What is the position at 9 seconds? 1. 29.0 2. 23.5 3. 31.5 4. 30.0 5. 21.0 6. 19.5 7. 28.0 8. 41.5 9. 44.0 10. 24.0 Correct answer: 24 m. Explanation: b b b b b b b b 1 2 3 4 5 6 7 8 9 1 2 3 4 − 1 − 2 time (s) velocity(m/s) The acceleration during the first 2 seconds is a = Δ v Δ t = 3 m / s − 0 m / s 2 s − 0 s = 1 . 5 m / s 2 . The position at 2 seconds is 10 m plus the area of the triangle (shaded above) x = 10 m + 1 2 (2 s − 0 s)(3 m / s − 0 m / s) = 13 m ; however, it can also be calculated: x = x i + v i ( t f − t i ) + 1 2 a ( t f − t i ) 2 = 10 m + (0 m / s) (2 s − 0 s) + 1 2 (1 . 5 m / s 2 )(2 s − 0 s) 2 = 13 m . The acceleration during the second time interval is a = Δ v Δ t = 4 m / s − 3 m / s 6 s − 2 s = 0 . 25 m / s 2 . The position at 6 seconds is 13 m plus the area of the trapezoid x = 13 m + 1 2 (6 s − 2 s)(4 m / s + 3 m / s) = 27 m ; however it can also be calculated: x = x i + v i ( t f − t i ) + 1 2 a ( t f − t i ) 2 = 13 m + (3 m / s) (6 s − 2 s) + 1 2 (0 . 25 m / s 2 )(6 s − 2 s) 2 = 27 m ....
View Full Document

This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 13

md1 - Version 072/ABACA – midterm 01 – Turner –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online