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HW2-phy2048-Spring-2014-solution

# The time at which the bullet strikes the target is t

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Unformatted text preview: s)2 = 12.0 m. 2 2 (b) The horizontal component of the velocity when it strikes the wall does not change from its initial value: vx = v0 cos 40.0° = 19.2 m/s. (c) The vertical component becomes (using Eq. 4- 23) vy = v0 sin θ0 − gt = (25.0 m/s) sin 40.0° − (9.80 m/s 2 )(1.15 s) = 4.80 m/s. 4P35. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4- 22 are directly applicable. The coordinate origin is at the end of the rifle (the initial point for the bullet as it begins projectile motion in the sense of § 4- 5), and we let θ0 be the firing angle. If the target is a distance d away, then its c...
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