For a projectile launched with an initial speed v 0 and angle theta above horizo

Question

For a projectile launched with an initial speed v 0 and angle theta above horizontal, the x and y components of the initial velocity are: v 0x = v0 costheta V 0y = v 0sin theta Due to the constant downward acceleration of gravity (g), the equations of motion are Range = x - x 0 = v 0x t y - y0 = v 0y t - 1/2 gt 2 where x0 is the starting horizontal position. y0 is the starting height x is the final horizontal position, y is the final height, t is the total time-of-flight, and g is the acceleration of gravity. As the final height equals the initial height (y = y 0) and equation (4) can be easily solved for t: t = 2v oy/g Using equation (2) to substitute for v 0y yields: t = 2v0/g sin theta Time-of-flight vs. launching angle for one-level projectile motion. Substituting equations (5) and (1) into equation (3) yields: Range= x-x 0 = 2v 0 2/g sin theta cos theta = v0 2/g sin 2theta Range vs. launching angle for one-level projectile motion. The initial velocity will be determined from your data. You will have two values for initial velocity, one from the time-of-flight data and one from the range data.

Explanation / Answer

---time of ascent=t at the maximum height the vertical component of the velocity=0 initial component in the vertical direction=vosin so using v=u-gt 0=vosin-gt so t=vosin/g when there is no air resistance time of ascent=time of descent hence time of descent=vosin/g the total time of flight=2vosin/g------5 -------------------------------------------- range=horizontal component of velocity *time of flight vocos*2vosin/g but 2sincos=sin2 hence range=x-xo=vo22sincos/g=vo2sin2/g------6 range will be maximum at 45 degree. time will be maximum when =90 degree.

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