A water fountain shoots a stream of water at an angle of 60^ to the horizontal. If the maximum height reached by the water droplets is 15 m , what is the initial speed of the water? (Take g = 10 m s ⁻² )

Step-by-step solution

The maximum height H of a projectile is given by the formula: H = u^2 ^2 2g Given that H = 15 m , = 60^ , and g = 10 m s ⁻² , we can substitute these values into the formula: 15 = u^2 ^2(60^ ) 2 10 Since (60^ ) = 3 2 , we have ^2(60^ ) = 3 4 . 15 = u^2 3 4 20 15 20 = u^2 3 4 300 = u^2 3 4 u^2 = 300 4 3 = 400 u = 400 = 20 m s ⁻¹ Answer: 20 m s ⁻¹