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?

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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}}

Answer

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}}

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?

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