A catapult launches a watermelon from a height of $80 \mathrm{~m}$ to a target located * 4 poinis on the ground. The watermelon leaves the catapult moving at $100 \mathrm{~km} / \mathrm{h}$ at an angle of $40^{\circ}$. How long is the watermelon in the air for?

Step 1 ：Calculate the horizontal and vertical components of the initial velocity: \(v_x = v * \cos(\theta)\) and \(v_y = v * \sin(\theta)\) where \(v = 27.78 \mathrm{~m/s}\) and \(\theta = 40^\circ\)

Step 2 ：Use the vertical motion equation to find the time in the air: \(0 = v_y * t - \frac{1}{2} * g * t^2\) where \(v_y = 17.86 \mathrm{~m/s}\) and \(g = 9.81 \mathrm{~m/s^2}\)

Step 3 ：\boxed{t \approx 6.25 \mathrm{~seconds}}