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A basketball player who is 2.0m tall is standing on the floor 10.0m from the basket,If he shoots the ball at a 40.0¤ angle witi the horizontal,at what initial speed must he throw so that it goes through the hoop without striking the backboard?The basket height is 3.05 m.

  • physical -

    consider the vertical
    hf=hi+Vsin40*t-4.9 t^2

    consider the horrizontal
    10=Vcos40 t
    V=10/cos40t

    put that in the first equation for V, then solve for t (notice it is a bit of algebra). have a pad of paper handy.

    Then after you have t, V=10/tcos40

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