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Natural Science Forum / Physics / General Physics / July 2008An easy trajectory problem (or is it) ?
If a bullet is fired at a 20 degree angle and lands 750 yards away. How far would it have traveled if it was fired at a 45 degree angle? (please ignore wind, humidity, temperature, air pressure, friction, spin.... etc.) | If a bullet is fired at a 20 degree angle and lands 750 yards away. | How far would it have traveled if it was fired at a 45 degree angle? | (please ignore wind, humidity, temperature, air pressure, friction, | spin.... etc.) Insufficient data. Velocity unknown, angle unknown, height unknown. (20 degrees from vertical is 70 degrees from horizontal) You've shot yourself in the foot at 20 degrees from vertical because you fired from the hip. > | If a bullet is fired at a 20 degree angle and lands 750 yards away. > | How far would it have traveled if it was fired at a 45 degree angle? [quoted text clipped - 5 lines] > You've shot yourself in the foot at 20 degrees from vertical because > you fired from the hip. Insufficient answer. Velocity can be determined from the angle and range. Angle was stated (20 degrees - from the HORIZONTAL, not VERTICAL). Height can be determined from the velocity and the angle. On Jul 12, 1:52 am, "Androcles" <Headmas...@Hogwarts.physics> wrote: > "When You Need to Tell Someone" <photogra...@yahoo.com> wrote in > messagenews:ed589d64-79f3-40e3-8190-c941fd060317@p25g2000hsf.googlegroups.com... [quoted text clipped - 7 lines] > You've shot yourself in the foot at 20 degrees from vertical because > you fired from the hip. Insufficient answer. Velocity can be determined from the angle and range. Angle was stated (20 degrees - from the HORIZONTAL, not VERTICAL). ------------------------------------------------------------------------------------- To travel 750 yards with an initial upward trajectory the bullet would be low velocity from an air rifle or pop gun and follow a markedly curved path as it fell under gravity. To travel 750 yards with an intial downward trajectory the bullet would be high velocity fired from a rifle in a building and have an *almost* straight line path. Insufficienct data, angle not known. ============================================= On Jul 11, 5:44 pm, When You Need to Tell Someone <photogra...@yahoo.com> wrote: > If a bullet is fired at a 20 degree angle and lands 750 yards away. > How far would it have traveled if it was fired at a 45 degree angle? > (please ignore wind, humidity, temperature, air pressure, friction, > spin.... etc.) Obvious homework question. Some hints: "parabola" "muzzle energy" "mgh". If you cannot Google your answer using those hints, consider a career in management. Mark L. Fergerson On Jul 12, 2:30 pm, "n...@bid.nes" <Alien8...@gmail.com> wrote: > On Jul 11, 5:44 pm, When You Need to Tell Someone > [quoted text clipped - 12 lines] > > Mark L. Fergerson Be careful what you call "obvious". I'm doing research for a book and I haven't had homework assignments for nearly 20 years (after receiving my MBA) On Jul 12, 8:17 pm, When You Need to Tell Someone <photogra...@yahoo.com> wrote: > On Jul 12, 2:30 pm, "n...@bid.nes" <Alien8...@gmail.com> wrote: > [quoted text clipped - 18 lines] > I haven't had homework assignments for nearly 20 years (after > receiving my MBA) Then you already have a career in management. That explains a lot. Mark L. Fergerson On Jul 11, 8:44 pm, When You Need to Tell Someone <photogra...@yahoo.com> wrote: > If a bullet is fired at a 20 degree angle and lands 750 yards away. > How far would it have traveled if it was fired at a 45 degree angle? > (please ignore wind, humidity, temperature, air pressure, friction, > spin.... etc.) theta = 20 deg = ~0.349 rad Deltax = 750 yd. = ~686 m theta' = 45 deg = ~0.785 rad Deltax = (a_x*Deltat^2)/2 + v_x_0*Deltat + x_0 = u_0*cos(theta)*Deltat = ~686 m a_x = 0, x_0 = 0, v_x_0 = u_0*cos(theta) Deltay = (a_y*Deltat^2)/2 + v_y_0*Deltat + y_0 = (g_0*Deltat^2)/2 + u_0*sin(theta)*Deltat = 0 y_0 = 0, v_y_0 = u_0*sin(theta), a_y = g = g_0 = ~-9.81 m/s^2 Deltat = (~0.0698 s^2/m)u_0 u_0 = (~730 m)/Deltat = (~10,500 m^2/s^2)/u_0 = ~102 m/s u_0^2 = ~10,500 m^2/s^2 Deltat' = -2sin(theta')*u_0/g_0 = ~14.8 s Deltax' = u_0*cos(theta')*Deltat' = ~1,070 m = ~1,170 yd. On Jul 11, 8:44 pm, When You Need to Tell Someone <photogra...@yahoo.com> wrote: > If a bullet is fired at a 20 degree angle and lands 750 yards away. > How far would it have traveled if it was fired at a 45 degree angle? > (please ignore wind, humidity, temperature, air pressure, friction, > spin.... etc.) theta = 20 deg = ~0.349 rad Deltax = 750 yd. = ~686 m theta' = 45 deg = ~0.785 rad Deltax = (a_x*Deltat^2)/2 + v_x_0*Deltat + x_0 = u_0*cos(theta)*Deltat = ~686 m a_x = 0, x_0 = 0, v_x_0 = u_0*cos(theta) Deltay = (a_y*Deltat^2)/2 + v_y_0*Deltat + y_0 = (g_0*Deltat^2)/2 + u_0*sin(theta)*Deltat = 0 y_0 = 0, v_y_0 = u_0*sin(theta), a_y = g = g_0 = ~-9.81 m/s^2 Deltat = (~0.0698 s^2/m)u_0 u_0 = (~730 m)/Deltat = (~10,500 m^2/s^2)/u_0 = ~102 m/s u_0^2 = ~10,500 m^2/s^2 Deltat' = -2sin(theta')*u_0/g_0 = ~14.8 s Deltax' = u_0*cos(theta')*Deltat' = ~1,070 m = ~1,170 yd. ================================================ If the 20 degrees is downward from a tall building or balloon, 45 degrees brings the bullet to LESS than 750 yards. > On Jul 11, 8:44 pm, When You Need to Tell Someone > [quoted text clipped - 18 lines] > Deltat' = -2sin(theta')*u_0/g_0 = ~14.8 s > Deltax' = u_0*cos(theta')*Deltat' = ~1,070 m = ~1,170 yd. Thank you. It's just what the doctor ordered. Now I can sleep well tonight. I was beginning to think that "not enough information" was a correct statement. Apparently, it wasn't. On Jul 13, 1:28 am, Zhividya <nsa...@gmail.com> wrote: > On Jul 11, 8:44 pm, When You Need to Tell Someone > [quoted text clipped - 18 lines] > Deltat' = -2sin(theta')*u_0/g_0 = ~14.8 s > Deltax' = u_0*cos(theta')*Deltat' = ~1,070 m = ~1,170 yd. Thank you. It's just what the doctor ordered. Now I can sleep well tonight. I was beginning to think that "not enough information" was a correct statement. Apparently, it wasn't. =============================================== It certainly was. Just because the fool assumed the angle was up and not down doesn't mean you gave enough information. Get another doctor, that cough might be TB even if you assumed it was just a common cold and the quack agreed with you. Oh, and sleep well. |
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