Answer:
The velocity is [tex]u = 19.1 \ m/s[/tex] Â
Explanation:
From the question we are told that
 The time taken by the ball is  t =  2.5 seconds
 The angle made with the horizontal is [tex]\theta = 40^o[/tex]
Generally the vertical component of the initial velocity is mathematically represented as
     [tex]u_y = u sin (\theta )[/tex]
Generally at maximum height of the ball the velocity is  v  =  0 m/s and  the time take to reach this maximum height is
  =>     [tex]t_m = \frac{t}{2} = \frac{2.5}{2} = 1.245 \ s[/tex]  [This because t is duration for the to a fro travel of the ball]
Generally from kinematic equation
    [tex]v = u_y + gt[/tex]
Here  [tex]g = -9.8 m/s^2[/tex] given that the direction is against gravity
=> Â Â [tex]0 = u sin (40) - 9.8 * [1.25][/tex]
=> Â Â [tex]0 = 0.6428u - 12.25[/tex] Â
=> Â Â [tex]u = 19.1 \ m/s[/tex] Â
