HW2-phy2048-Spring-2014-solution

8 ms 2 v0 x 10 m 157 ms 2y 2 20 m top

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Unformatted text preview: y = h > 0 at t = 1.50 s, we have 1 y − y0 = v0 y t − 2 gt 2 where v0y = v0 sin 60.0°. This leads to h = 32.3 m. (b) We have vx = v0x = (33.3 m/s)cos 60.0° = 16.7 m/s vy = v0y – gt = (33.3 m/s)sin 60.0° – (9.80 m/s2)(1.50 s) = 14.2 m/s. r The magnitude of v is given by ! 2 | v |= vx + v 2 = (1 6 . 7 m / s)2 + (1 4 . 2 m / s)2 = 2 1 . 9 m / s . y (c) The angle is ⎛ vy ⎞ − 1 ⎛ 14.2 m/s ⎞ ⎟ = tan ⎜ ⎟ = 40.4 ° . ⎝ 16.7 m/s ⎠ ⎝ vx ⎠ θ = tan −1 ⎜ 4 PHY2048 Spring 2014 HW2 4P67. The stone moves in a circular path (top view shown b...
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