#$&* course Phy 241 November 9 around 1pm. 023. `query 23*********************************************
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Given Solution: `a** We know the following: For y motion `dsy = + 8 m, ay = -g = - 9.8 m/s^2 and v0y = v0 sin(60 deg) = .87 v0. For x motion `dsx = 18 m, ax = 0 and v0x = v0 cos(60 deg) = .5. Assuming a coordinate system where motion starts at the origin: The equation of motion in the x direction is thus x = .5 v0 * t and the equation of y motion is y = .87 v0 t - .5 g t^2. We know x and y at impact and we know g so we could solve these two equations simultaneously for v0 and t. We begin by eliminating t from the two equations: x = .5 v0 * t so t = 2 x / v0. Substituting this expression for t in the second equation we obtain y = .87 v0 * (2 x / v0) - .5 g ( 2 x / v0) ^ 2. Multiplying both sides by v0^2 we obtain v0^2 y = .87 v0^2 ( 2 x) - .5 g * 4 x^2. Bringing all the v0 terms to the left-hand side we have v0^2 y - 1.73 v0^2 x = -2 g * x^2. Factoring v0 we have v0^2 ( y - 1.73 x) = -2 g x^2 so that v0 = +-sqrt(-2 g x^2 / ( y - 1.73 x) ) ) . Since we know that y = 8 m when x = 18 m we obtain = +- sqrt( -9.8 m/s^2 * (18 m)^2 / ( 8 m - 1.73 * 18 m) ) = +-sqrt(277 m^2 / s^2) = +-16.7 m/s, approx.. We choose the positive value of v0, since the negative value would have the projectile moving 'backward' from its starting point. Substituting this value into t = 2 x / v0 and recalling that our solution applies to the instant of impact when x = 16 m and y = 8 m we obtain t = 2 * 18 m / (16.7 m/s) = 2.16 s. Alternatively we can solve the system for v0 less symbolically and perhaps gain different insight into the meaning of the solution. Starting with the equations x = .5 v0 * t and y = .87 v0 t - .5 g t^2 we see that impact occurs when x = .5 v0 t = 18 m so that t = 18 m / (.5 v0) = 36 m / v0. At this instant of impact y = 8 m so substituting this and the t just obtained into the equation of motion for y we get y = .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m. The equation .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m is easily solved for v0, obtaining v0 = 16.7 m/s. With this initial velocity we again confirm that t = 2.16 sec at impact. Note that at t = 2.16 sec we get y = 14.4 m/s * 2.16 s - 4.9 m/s^2 * (2.16 s)^2 = 8 m, within roundoff error, confirming this solution. We need the magnitude and direction of the velocity at impact. We therefore need the x and y components of the velocity at the t = 2.16 sec instant. At this instant we have x and y velocities vx = dx/dt = .5 v0 = 8.35 m/s and vy = dy/dt = 14.4 m/s - 9.8 m/s^2 * t = 14.4 m/s - 9.8 m/s^2 * 2.16 s = -5.6 m/s, approx. The velocity at impact therefore has magnitude sqrt( (8.25 m/s)^2 + (-5.6 m/s)^2 ) = 10 m/s, approx. and the angle is arctan(vy/vx) = arctan(-5.6 / 8.25) = -34 deg, approx. At impact the object is moving at 10 m/s and at 34 deg below horizontal (i.e., it's on its way back down). ** STUDENT COMMENT I think I have the right understanding of how to do this but I keep getting different answers than the given ones. They seem to use many rounding errors. For example, from the given solution above I found this vy = dy/dt = 14.4 m/s - 9.8 m/s^2 * t = 14.4 m/s - 9.8 m/s^2 * 2.16 s = -5.6 m/s, approx. When doing this in my calculator, I get -6.7689, so I don't see how they think -5.6 is approximate. INSTRUCTOR RESPONSE looks like I approximated very roughly, using 9.8 * 2.16 = 20. Clearly 9.8 * 2.16 is closer to 21. Should have been more careful since the result was then used in a subtraction. Remember, though, that my numbers are mostly the result of mental approximation. That is by design; they aren't intended to be all that accurate, though in most cases they should be reasonably close. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Some different answers, probably just due to rounding and sig fig errors. ------------------------------------------------ self-critique rating #$&*: