grade 11 physics

A championship golfer uses a nine iron to chip a shot right
into the cup. If the golf ball is launched at a velocity of
20 m/s at an angle of 45° above the horizontal, how far
away was the golfer from the hole when he hit the ball?
What maximum height did the ball reach?

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  1. Range = (Vo^2/g)*sin2A
    In this case, A = 45 degrees, so
    Range = Vo^2/g
    (This assumes the ball does not roll in to the cup))

    The maximum height, when hit at A = 45 degrees, is such that
    H = (Vo sinA)^2/g
    = Vo^2/(2g) if A = 45 degrees

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