v_o = initial velocity of lion

= angle of launch

consider the motion along the vertical direction or Y-direction from initial position to the point of maximum height.

V_oy = initial velocity of lion In Y-direction = V_o Sin

a = acceleration = - 9.8 m/s²

Y_max = maximum height gained = 3 m

V_fy = final velocity in Y-direction at the maximum height = 0 m/s

using the equation

V_fy² = V_oy² + 2 a Y_max

inserting the values

0² = V_oy² + 2 (- 9.8) (3)

V_oy² = 58.8

V_oy = sqrt(58.8)

V_oy = 7.7 m/s

V_o Sin = 7.7 m/s eq-1

consider the motion along the vertical direction or Y-direction from initial to final position.

V_oy = initial velocity of lion In Y-direction = V_o Sin

a = acceleration = - 9.8 m/s²

Y = displacement = 0 m

t = time of travel

using the equation

Y = V_oy t + (0.5) at²

0 = (V_o Sin) t + (0.5) (- 9.8) t²

using eq-1

0 = (7.7) t + (0.5) (- 9.8) t²

t = 1.6 sec

consider the motion along the horizontal direction or X-direction

V_ox = initial velocity In X-direction = V_o Cos

a = acceleration = 0

X = displacement = 10

t = time of travel = 1.6

using the equation

X = V_ox t + (0.5) at²

10 = (V_o Cos ) (1.6)+ (0.5) (0)(1.6)²

V_o Cos = 6.3 m/s eq-2

dividing eq-1 by eq-2

V_o Sin/(V_o Cos) = 7.7/6.3

tan = 1.22

 = tan^-1(1.22)

 = 50.7 deg

using eq-1

V_o Sin = 7.7

inserting the value of the angle

V_o Sin50.7 = 7.7

V_o = 9.95 m/s

please see the attached file for the complete solution