An object is launched with an initial velocity of 50.0 {\ m m/s} at a launch ang

Question

An object is launched with an initial velocity of 50.0 { m m/s} at a launch angle of 36.9^circ above the horizontal.
Determine the speed v at each 1 { m s} from t = 0 { m s} to t = 6 { m s}

Explanation / Answer

Initial vertical velocity = 50*SIN(Radians(36.9)) = 30m/sec Initial Horizontal velocity = 50*COS(Radians(36.9)) = 40m/sec Horizontal velocity remains constant as there is no acceleration or deceleration in horizontal direction. Vertical Direction: At t = 0sec, v = 30-((9.81)*0) = 30m/sec => Speed = 50m/sec At t = 1sec, v = 30-((9.81)*1) = 20.19m/sec => Speed = 44.80665241m/sec At t = 2sec, v = 30-((9.81)*2) = 10.38m/sec => Speed = 41.32486419m/sec At t = 3sec, v = 30-((9.81)*3) = 0.57m/sec => Speed = 40.00406104m/sec At t = 4sec, v = 30-((9.81)*4) = -9.24m/sec => Speed = 41.05335065m/sec At t = 5sec, v = 30-((9.81)*5) = -19.05m/sec => Speed = 44.30465551m/sec At t = 6sec, v = 30-((9.81)*6) = -28.86m/sec => Speed = 49.32443208m/sec

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