52. For Δ y = 0, Eq. 4-22 leads to t = 2v o sin θ o / g , which immediately implies t max = 2 v o / g (which occurs for the “straight up” case: o = 90 ° ). Thus, 1 2 t max = v o / g ¡ 1 2 = sin o . Therefore, the half-maximum-time flight is at angle o = 30.0 ° . Since the least speed occurs at the top of the trajectory, which is where the velocity is simply the x-component of the initial velocity (
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.