Asked by yelya
                A ball launched from ground level lands 2.15 s later on a level field 50 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)
            
            
        Answers
                    Answered by
            Damon
            
    u = constant horizontal speed = 50/2.5 = 20 m/s
v = vi - 9.8 t
at top of arc, t = 2.5/2 = 1.25 seconds and v = 0
so at top
0 = vi - 9.8 (1.25)
so
vi = 12.25
so speed at start = sqrt (20^2 + 12.25^2)
= sqrt(400+ 150) = sqrt 550 = 23.5 m/s
tan angle = 12.25/20
so
angle = 31.5 degrees
    
v = vi - 9.8 t
at top of arc, t = 2.5/2 = 1.25 seconds and v = 0
so at top
0 = vi - 9.8 (1.25)
so
vi = 12.25
so speed at start = sqrt (20^2 + 12.25^2)
= sqrt(400+ 150) = sqrt 550 = 23.5 m/s
tan angle = 12.25/20
so
angle = 31.5 degrees
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