Physics

A ball thrown by a boy in the street is caught 5 s later by another boy on the balcony of the house 10 m away and 3 m above the street level. What is the initial speed of the ball? What is the angle above the horizontal at which it was thrown?

asked by Kadu
  1. If it takes 5 seconds to cover that distance you will have to throw it just about straight up so it will stay up five seconds !

    u = constant horizontal speed = 10m/5s
    = 2 meters/second

    in 5 seconds it goes up 3 meters

    3 = Vi t - 4.9 t^2
    3 = Vi(5) = 4.9(25)
    3 = 5 Vi - 122.5
    Vi = 25.1 m/s
    so
    speed = sqrt (25.1^2 + 2^2)
    = 25.2 m/s
    tan T = Vi/u = 25.2/2
    T = 85.5 degrees

    posted by Damon

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