Let's first find the time that the ball takes to reach the ground:
y = y 0 − v 0 t s i n θ − 1 2 g t 2 , y=y_0-v_0tsin\theta-\dfrac{1}{2}gt^2, y = y 0 − v 0 t s in θ − 2 1 g t 2 , 0 = 12 − 4.7 t ⋅ s i n 2 5 ∘ − 4.9 t 2 , 0=12-4.7t\cdot\ sin25^{\circ}-4.9t^2, 0 = 12 − 4.7 t ⋅ s in 2 5 ∘ − 4.9 t 2 , 4.9 t 2 + 1.98 t − 12 = 0. 4.9t^2+1.98t-12=0. 4.9 t 2 + 1.98 t − 12 = 0. This quadratic equation has two roots: t 1 = 1.37 s t_1=1.37\ s t 1 = 1.37 s t 2 = − 1.78 s t_2=-1.78\ s t 2 = − 1.78 s t = 1.37 s t=1.37\ s t = 1.37 s
The horizontal component of the ball's velocity doesn't change during the flight and can be found as follows:
v x = v 0 c o s θ = 4.7 m s ⋅ c o s 2 5 ∘ = 4.26 m s . v_x=v_0cos\theta=4.7\ \dfrac{m}{s}\cdot cos25^{\circ}=4.26\ \dfrac{m}{s}. v x = v 0 cos θ = 4.7 s m ⋅ cos 2 5 ∘ = 4.26 s m . The vertical component of the ball's velocity can be found as follows:
v y = v 0 s i n θ − g t , v_y=v_0sin\theta-gt, v y = v 0 s in θ − g t , v y = 4.7 m s ⋅ s i n 2 5 ∘ − 9.8 m s 2 ⋅ 1.37 s = − 11.44 m s . v_y=4.7\ \dfrac{m}{s}\cdot sin25^{\circ}-9.8\ \dfrac{m}{s^2}\cdot1.37\ s=-11.44\ \dfrac{m}{s}. v y = 4.7 s m ⋅ s in 2 5 ∘ − 9.8 s 2 m ⋅ 1.37 s = − 11.44 s m . Finally, we can find the speed of the ball when it reaches the ground from the Pythagorean theorem:
v = v x 2 + v y 2 = ( 4.26 m s ) 2 + ( − 11.44 m s ) 2 = 12.2 m s . v=\sqrt{v_x^2+v_y^2}=\sqrt{(4.26\ \dfrac{m}{s})^2+(-11.44\ \dfrac{m}{s})^2}=12.2\ \dfrac{m}{s}. v = v x 2 + v y 2 = ( 4.26 s m ) 2 + ( − 11.44 s m ) 2 = 12.2 s m . 
