A golf ball is struck by a club such that it leaves the ground with an initial velocity of 40 m s ⁻¹ at an angle of 60^ with the horizontal. Considering air resistance to be negligible, what is the maximum vertical height reached by the ball? (Take g = 10 m s ⁻² )

Step-by-step solution

Initial velocity, u = 40 m s ⁻¹ Angle of projection, = 60^ Acceleration due to gravity, g = 10 m s ⁻² The maximum vertical height H is given by the formula: H = u^2 ^2 2g Substituting the given values: H = (40)^2 ^2 60^ 2 10 H = 1600 ( 3 2 )^2 20 H = 1600 3 4 20 H = 1200 20 = 60 m Answer: 60 m