Posted by **sayan** on Tuesday, April 10, 2012 at 12:38am.

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?

- grade 11 physics -
**drwls**, Tuesday, April 10, 2012 at 1:45am
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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