{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

PHY303k_-_Spring_2010_-_HW_5

# PHY303k_-_Spring_2010_-_HW_5 - saldana(avs387 Homework 05...

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

saldana (avs387) – Homework 05 – florin – (58140) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The speed of an arrow fired from a compound bow is about 18 m / s. An archer sits astride his horse and launches an arrow into the air, elevating the bow at an angle of 67 above the horizontal and 1 . 5 m above the ground. The acceleration of gravity is 9 . 81 m / s 2 . 18 m / s 67 1 . 5 m range What is the arrow’s range? Assume: The ground is level. Ignore air resistance. Correct answer: 24 . 3785 m. Explanation: Let : v o = 18 m / s , θ = 67 , h = 1 . 5 m , and g = 9 . 81 m / s 2 . 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 sin 2 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 = (18 m / s) 4 (sin 2 134 ) + (8) (9 . 81 m / s 2 ) × (18 m / s) 2 (1 . 5 m) (cos 2 67 ) = 60142 . 9 m 2 / s 2 , then x = (18 m / s) 2 (sin 134 ) 2 (9 . 81 m / s 2 ) + radicalbig 60142 . 9 m 2 / s 2 2 (9 . 81 m / s 2 ) = 24 . 3785 m .

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

View Full Document
saldana (avs387) – Homework 05 – florin – (58140) 2 002 (part 2 of 2) 10.0 points Now assume his horse is at full gallop, moving in the same direction as he will fire the arrow, and that he elevates the bow the same way as before. What is the range of the arrow at this time if the horse’s speed is 21 m / s? Correct answer: 97 . 1692 m. Explanation: The arrow’s speed relative to the ground and the angle of elevation relative to the ground are changed v x = v arrow + v archer = (18 m / s) cos 67 + 21 m / s = 28 . 0332 m / s . v y = (18 m / s) sin 67 = 16 . 5691 m / s The arrow’s speed relative to the ground is then given by v o = radicalBig v 2 x + v 2 y = radicalBig (28 . 0332 m / s) 2 + (16 . 5691 m / s) 2 = 32 . 5637 m / s , and the angle of elevation relative to the ground is θ = tan 1 v y v x = tan 1 parenleftbigg 16 . 5691 m / s 28 . 0332 m / s parenrightbigg = 30 . 5853 . With the new speed v 0 and angle θ , the new range x can be found similarly to the previous part as x = v 2 o sin 2 θ 2 g ± 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 = (32 . 5637 m / s) 4 (sin 2 30 . 5853 ) +(8) (9 . 81 m / s 2 ) (32 . 5637 m / s) 2 × (1 . 5 m) (cos 2 30 . 5853 ) = 9 . 55492 × 10 5 m 2 / s 2 , then x = (32 . 5637 m / s) 2 (sin 30 . 5853 ) 2 (9 . 81 m / s 2 ) + radicalbig 9 . 55492 × 10 5 m 2 / s 2 2 (9 . 81 m / s 2 ) = 97 . 1692 m .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

PHY303k_-_Spring_2010_-_HW_5 - saldana(avs387 Homework 05...

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

View Full Document
Ask a homework question - tutors are online