Question
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?
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?
Answers
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
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