Aboy kicks a football from ground level. the ball takes 3 seconds to reach its maximum height. what is the angle of the initial velocity with respect to the horizontal, if the ball lands 15 m away from the launching point?

Explanation:

Given that the ball takes 3 seconds to reach its maximum height.

Consider the vertical motion of ball till maximum height.

We have equation of motion v = u + at

     Acceleration, a = -9.81 m/s²

     Final velocity, v = 0 m/s    

     Time, t = 3 s

     Substituting

                      v = u + at  

                      0 = u + -9.81 x 3

                      u = 29.43 m/s

Initial vertical velocity is 29.43 m/s.

Now consider horizontal motion of ball.

Time of flight of ball = 2 x Time to reach maximum height = 2 x 3 = 6 s

Displacement = 15 m

We have equation of motion s = ut + 0.5 at²

        Displacement, s = 15 m

        Acceleration, a = 0 m/s²  

        Time, t = 6 s      

     Substituting

                      s = ut + 0.5 at²

                      15 = u x 6 + 0.5 x 0 x 6²

                      u = 2.5 m/s

Initial horizontal velocity is 2.5 m/s

Let r be the initial velocity and θ be the angle with horizontal

              Initial vertical velocity = rsinθ = 29.43 m/s

              Initial horizontal velocity = rcosθ = 2.5 m/s

Dividing

The angle of the initial velocity with respect to the horizontal is 85.14°

i believe the answer is a hope it and is correct