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Introductory Physics Homework Help
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[QUOTE="barthayn, post: 2594541, member: 195458"] [b]Are these right?[/b] Hi, I just did 2/3 of my assignment for physics and had trouble with the first question. I was wondering if I was correct. Please look over my work. [h2]Homework Statement [/h2] 3. Two planes fly from Toronto to Philadelphia. Plane A flies via Pittsburgh where as passengers on plane B have a direct flight. Pittsburgh is 350 hm due south of Toronto and 390 km due west of Philadelphia. The airspeed of both planes is 400km/h and a steady wind is blowing from the east at 60km/h. a)What direction must the pilot point the plane flying from Toronto to Pittsburgh. Include a vector diagram of velocities. b) How long will the entire flight take for plane A assuming a 0.50h layover in Pittsburgh? 4. A shell is fired from a cliff that is 36m above the horizontal plane. The muzzle speed of the shell is 80m/s and it is fired at an elevation of 25[sup]o[/sup] above the horizontal. a) Determine the horizontal range of the shell. b) Determine the velocity of the shell as it strikes the ground. [h2]Homework Equations[/h2] v = d/t y = ((v[sub]i[/sub]+v[sub]f[/sub])/2)t v[sub]f[/sub][sup]2[/sup] = v[sub]i[/sub][sup]2[/sup] + 2ay [h2]The Attempt at a Solution[/h2] 3. a) I drew a vector diagram. The hypothesis of the side of the triangle was 400km/h whilst the opposite was 60km/h. The other side (adjacent) turned out to be around 395km/h. This was a right angle triangle. I used TANθ=o/a TANθ = 60/395 TANθ = 0.1517 θ = 9[sup]o[/sup] Therefore the plane must travel 400km/h [9[sup]o[/sup] S of W] To reach Pittsburgh. Is this correct? b) I used the same diagram and got t = d/v t = 350/395 t = 0.885h V[sub]total[/sub] = v[sub]normal[/sub]-v[sub]air[/sub] V[sub]total[/sub] = 400 - 60 V[sub]total[/sub] = 340km/h t= d/v t= 390/340 t= 1.14 t[sub]total[/sub] = 1.14+.5+0.885 t[sub]total[/sub] = 2.5h Therefore it would take 2.5h with the 0.5h layover. 4 a) let up be negative v[sub]f[/sub][sup]2[/sup] = (-.33.8)[sup]2[/sup]+19.6*36 v[sub]f[/sub][sup]2[/sup] = 1143.08+705.6 v[sub]f[/sub][sup]2[/sup] = 1848.68 v[sub]f[/sub][sup]2[/sup] = 42.99m/s v[sub]f[/sub][sup]2[/sup] = 43m/s t = ((2y)/(v[sub]i[/sub]+v[sub]f[/sub])) t = 72 / 76.8057 t = 0.93s d = tv d = 80cos25*0.93 d = 68m Therefore the horizontal range of the shell is 68m. I used my final velocity for the y-axis to get two sides of the diagram because horizontal does not change. c[sup]2[/sup] = 42.99[sup]2[/sup] = (80cos25)[sup]2[/sup] c[sup]2[/sup] = 1848.68 c = 42.99m/s TANθ = o/a TANθ = (80cos25)/42.99 TANθ =1.6863 θ = 59.33 θ = 59 Therefore the shell hit the ground at 84m/s [59[sup]o[/sup] below the horizontal]. Are these all correct? Did I do 3 a correct? Where can I improve? [/QUOTE]
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