A golfer strikes a golf ball, which then moves off at an angle to the ground - Scottish Highers Physics - Question 3 - 2019

From the graph of vertical velocity $v_y$ versus time, we observe that just before the ball hits the ground, the vertical velocity reaches a maximum of $-30$ ms$^{-1}$ (the negative sign indicates downward direction). From the horizontal velocity $v_x$ graph, the horizontal component remains constant at $40$ ms$^{-1}$ throughout.

Using the Pythagorean theorem to calculate the resultant speed just before impact: $v = ext{sqrt}(v_x^2 + v_y^2)$ Substituting the values: $v = ext{sqrt}((40 ext{ ms}^{-1})^2 + (-30 ext{ ms}^{-1})^2)$ $v = ext{sqrt}(1600 + 900) = ext{sqrt}(2500) = 50 ext{ ms}^{-1}$