Adart is thrown at some speed v at 45° above the horizontal, with the aim of hitting the bull's eye of a dartboard located 2.5 m away. if the dart leaves the hand of the player at a height located 20 cm below the bull's eye, at what speed v must the dart be thrown?

let us say that t = time so that for horizontal motion:

t = distance / velocity * cos angle 
t = 2.5 / v cos 45 

while for the vertical motion, (s = height = 20 cm = 0.20m, t = time, g = gravity)
s = ut - 0.5 gt^2 
0.2 = v sin 45 * 2.5 / v cos 45 - 0.5 * 9.81 * [2.5 / v cos 45] ^2 
0.2= 2.5 tan 45 - 61.31 / v^2 
61.31 / v^2 = 2.5 - 0.2 = 2.3 
v = 5.16 m/s 

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.