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80 ms 2 115 s2 120 m 2 2 b the horizontal component

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Unformatted text preview: he positive direction choices used in the textbook so that equations such as Eq. 4- 22 are directly applicable. The coordinate origin is at the release point (the initial position for the ball as it begins projectile motion in the sense of §4- 5), and we let θ0 be the angle of throw (shown in the figure). Since the horizontal component of the velocity of the ball is vx = v0 cos 40.0°, the time it takes for the ball to hit the wall is 2 PHY2048 Spring 2014 t= HW2 Δx 22.0 m = = 1.15 s. vx (25.0 m/s) cos 40.0° (a) The vertical distance is 1 1 Δy = (v0 sin θ0 )t − gt 2 = (25.0 m/s)sin 40.0°(1.15 s) − (9.80 m/s 2 )(1.15...
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