A ball is projected at a speed of 40 m/s at an angle of 60^ to the horizontal. Determine the maximum height reached by the ball in m. Take g = 10 m/s ^2 .

Step-by-step solution

The maximum height H reached by a projectile is given by the formula: H = u^2 ^2 2g Given u = 40 m/s , = 60^ , and g = 10 m/s ^2 . Substituting the values: H = (40)^2 ( 60^ )^2 2 10 H = 1600 ( 3 2 )^2 20 H = 1600 3 4 20 H = 1200 20 = 60 m