grade 11 physics
posted by sayan .
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?
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