{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Sol-Ch04 - CHAPTER 4 E xercises 1(a v = 6t 2 i 3 t 2 j = 1...

This preview shows pages 1–4. Sign up to view the full content.

Exercises 1. (a) v = (6t – 2) i – 3t 2 j   = 10i – 12 j   m / s; (b) a = 6i – 6tj = 6i – 24 j   m / s 2 , (c) a a v = 6i –12 j   m / s 2 . 2. (a) 0.5i – 1.5 j   + 2k m / s; (b) v 1 = v 2 – a Δ t = 40i – 10 j   – 2k m / s 3. (a) v a v = 4i + 7 j   m / s ; (b) v i = 12i + 9 j   and v f = 18i + 24 j     ; so a a v = 1.2i + 3 j   m / s 2 . 4. (a) 350 m; (b) 250 m, 37° N of E; (c) 10 m / s, 37° N of E; (d) 14 m / s ; (e) a av = (10 j   – 20i ) / 25 = –0.8i + 0.4 j   m / s 2 5. (a) Δ r = (0.707Ri + 0.707R j   ) – R j   = –09i – 0.37 j   m; (b) Δ r = (0.707Ri – 0.707R j   ) – R j   = 0.3i – 0.71 j   m / s; (c) Δ v = –(2) 1/2 i , a av = Δ v / Δ t = –0.71i m / s 2 6. The 5810 km distance is along the curved surface of the earth. The angular displacement is Δ θ = 5810 km / 6370 km = 0.912 rad. The magnitude of the displacement is the length of the straight line joining the initial and final points :2 tan(Δ θ / 2) = 5611 km. Thus, v av = 5611 km / 33.5 h = 167 km / h. 7. (a) From E q. 4.11, 0 =100 + 20t – 4.9t 2 , thus t = 7 s. (b) From E q. 4.12, 0 = 400 – 2gH, so H = 120 m (c) R = v x t = 105 m. (d) v y = 20 – g t = –48.6 m / s, so v = 15i – 48.6 j   m / s 8. (a) From E q. 4.11: 0 = 40 – (v 0 sin 37°) 2 – 4.9(2 2 ), so v 0 =16.9 m/s. Then, R = (v 0 cos 37°)t = 27.0 m; (b) v y = –v 0 sin 37° – 9.8t = –29.8 m / s, v x = 13.5 m / s. tan θ = v y / v x = 2.21, thus θ = 65.6° below the horizontal 9. From Example 4.2, R = v 0 2 sin 2 θ / g = 20.3 m From E q. 4.11: 0 = 0 + (v 0 cos 45°)t – 4.9t 2 , so t = 2.03 s. Person must run at v = (30 m – 20.3 m) / 2.03 s = 4.77 m / s toward the thrower. 10. From Example 4.2 : sin 2 θ = R g / v 0 2 thus θ = 31.4° or 58.6° CHAPTER 4

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

View Full Document
11. (a) t = (18.3 m) / (30 m / s) = 0.61 s, so Δ y = –4.9 t 2 = –1.82 m (b) x = 18.3 = 30 cos θ t ; y = 0 = 30 sin θ t – 4.9t 2 . Substitute t = 18.3 / (30 cos θ ) into the equation for y to find : Sin 2 θ = 9.8 (0.61) 2 /18.3 Thus θ = 5.75° or 84.2° (reject). D = 18.3 tan 5.75° = 1.84 m OR: From Example 4.2, sin 2 θ = R g / v 0 2 = 0.499, so θ = 5.75° 12. (a) From x = 3 = v 0 t and Δ y = –0.05 = –4.9t 2 , find v 0 = 29.7 m/s; (b) From Example 4.2, sin 2 θ = R g / v 0 2 = 0.0333, so θ = 0.95° 13. From Example 4.2, R = v 0 2 sin 2 θ / g, we find v 0 = 31.3 m / s From E q. 4.12, H = v 0 2 / 2g = 50 m 14. (a) 0 = 200 – v 0 sin 37°(4) – 4.9(4 2 ), so v 0 = 50.5 m / s ; (b) x = v 0 cos 37° t = 161 m 15. (a) From v 0 2 = R g / sin 2 θ , we find v 0 = 9.81 m / s. (b) 10.8 m/s. 16. x = v 0 t, Δ y = –10 4 m = –4.9t 2 , so x = v 0 (10 4 / 4.9) 1/2 = 13.6 km 17. x = 4 = v 0 cos θ t; y = 0.8 = v 0 sin θ t – 4.9t 2 . Substitute t = 4 / (v 0 cos θ ) into the second equation to find v 0 = 7 m/s. 18. Use R = v 0 2 sin 2 θ / g with θ = 15° to find v 0 = 34.3 m / s 19. (a) 50 = 27 cos(32°)t, so t =2.18 s, then y = 1 + 27 sin(32°) – 4.9t 2 = 8.9 m. (b) R = v 0 2 sin 2 θ / g = 66.0 m, and the time of flight is t = R / (v 0 cos θ ) = 2.92 s. Player has Δ t = 2.92 – 0.5 = 2.42 s to run 16.9 m, so v = 16.9 m / 2.42 s= 7 m / s. 20. 1.6 = v 0 t, and –1 = –4.9t 2 , so t = 0.452 s and v 0 = 3.54 m / s 21. (a) Speed of rocket at 6.5 s is 52 m / s and it has risen to a height of (26 m / s)(6.5) sin70° = 169 sin 70° = 159 m. The rest of the trip is in free-fall. To find the highest point, 0 = (52 sin 70°) 2 – 2g Δy, yields Δ y = 122 m.
Thus H = 122 + 159 = 281 m. (b) In the initial phase the horizontal displacement is Δx 1 = 169 cos 70° = 57.8 m. During free-fall : X = 52 cos θ t; y = 0 = 159 + 52 sin θ t – 4.9t 2 , Which yields t = 12.55 s and then Δ x 2 = 223.2 m. The total horizontal range is R = 223.2 + 57.8 = 281 m.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}