Question

FULL SCREEN PRINTER Chapter 2, Practice Problem 2/040 A horseshoe player releases the horseshoe at A with an initial speed Vo = 36 ft/sec. Determine the ranges for the launch angle for which the shoe will strike the 16-in, vertical stake. 36 ft/sec 13.9 Answers: Lower range: sos Higher range:

Answer 1

The detailed solution is given below:

Given; Vo = 35ft/sec Set the x-y coordinates at Point A : -→ X-Motion: Ve 1 (Volx = Vo Cose , KA=0, X8 = 40ft ko ² xq = XA+ (volext (*) 40 = 0 + (36 x Coso )xt or Jose xt = → Y-motion: (Voly - Vo smo, 4x=0, 48 = $conteft top of the stake) (-3.4ft (bottom of the stake) and ag= mg = -322 » Yo = Ya + (vody xz + Į ayxt? 4 = 0 H36xsino ) Xt + (-32.2) x+? or Yo = 365ho xt -10.147 - and E-9 = -32.2 ft 2 18

par - 3.9x12 = 46.8 in L 46.8-16= 30.8 in /stake -YA (6เท bottom From , we get Put the value of t in @ we get - 16.1X /10 JB = 36 sino x 10 9x coso (98(050) - 40 tano - 16. 1 X 100 81 Cos²o Oy 4. Y = 40 tano - 19:8765 seco = 40 tono – 19,8765 (It tone) Coittan? O - sec 07

Yo = –14.8765 dando tAorton 0 – 19.8765 - © Now, put Je = -3.8 and Jp = -3.9 in -30,8 = -19.8765 tono + Anton 0–195765 — ® and –3.9= -19.8765 tone + to ton0 – 19.8765 — © → Divide ® by C-195765) -2.567 = tanpo +6–2-012) tamo + - 19.8765 or tan o -2.012 tano +/-0.129 = 0 or i tamo-2.012 tam 0 +0.871 = 0 Now fono = -(-2.012) + J(-.01272 – 4(1)(0571) 2x1 - 2012 +0.751 7 tano = 2.012+0.75) = 1-3815 | tano – 2.012-0.75) or 10 = 54.te] 54. 10° = 0.6305 OF 37.2307

→ Bride ☺aby (-19.5765) = tano+(-2.012) tanot -3. g -19.8765 or Hano - 2.012 tan 0 +0.804 - 0 - Now, tano= -62.012) ± √(-20012) ²4 (1) (0.804) 2X1 2.012 + 0.912 = 0 - 2.01270.912 tan o = 2.012-0.912 = 0.55 = 1.462 or 10 = 55.63") or 0 = 28.81° Ranges :- → Lower : 28.81' < 0% 32.23° Higher 54.10 <0 < 55.630