(a) since the range R = v^2/g sin2θ, we have
v^2/9.81 sin60° = 228
v = 50.82 m/s
(b) h = v^2/2g sin^2θ = 50.82^2/19.62 sin^2(30°) = 32.91 m
A golf ball with an initial angle of 30° lands exactly 228 m down the range on a level course.
(a) Neglecting air friction, what initial speed would achieve this result?
m/s
(b) Using the speed determined in item (a), find the maximum height reached by the ball.
2 answers
Thank you super helpful I understand now