HW2-phy2048-Spring-2014-solution

We see in this time reversed situation that it is

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Unformatted text preview: oordinates are x = d, y = 0. The projectile motion equations lead to d = (v0 cos θ 0 )t 0 = v0t sin θ 0 − 1 gt 2 2 The setup of the problem is shown in the figure. The time at which the bullet strikes the target is t = d /(v0 cos θ0 ) . Eliminating t leads to 2 2v0 sin θ 0 cosθ 0 − gd = 0. Using sin θ 0 cosθ 0 = 1 sin ( 2θ 0 ) , we obtain 2 2 v0 sin (2θ0 ) = gd ⇒ sin(2θ0 ) = gd (9.80 m/s 2 )(45.7 m) = 2 v0 (460 m/s) 2 3 PHY2048 Spring 2014 HW2 which yields sin(2θ 0 ) = 2.11×10 −3 , or θ0 = 0.0606°. If the gun is aimed at a point a distance ℓ above the target, then tan θ 0 = ℓ d so that l = d tan θ0 = (45.7 m) tan(0.0606°) = 0.0484 m = 4.84 cm....
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