Question

A water-skier with a mass of 68 kg is pulled with a constant force of 720 N bya speedboat. A wave launches him in such a way that he is temporarilyairborne while still being pulled by the boat, as shown in the image below.Assuming that air resistance can be ignored, what is the vertical accelerationthat the water-skier experiences on his return to the water surface? (Recallthat g = 9.8 m/s²)Rope Force720 N350WeightOA. -15.9 m/s²OB. -28.7 m/s²O C. -19.6 m/s²OD. -9.8 m/s²

Answer 1

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Step-by-step explanation:

Answer 2

A. -15.9 m/s²

Step-by-step explanation:

By the second law of newton, the net vertical force is equal to mass times vertical acceleration, so we can write the following equation:

`F_(net)=-F\sin35-mg=ma`

Where F is the constant force of 720 N, so Fsin(35) is the vertical component of this force, m is the mass, g is the gravity and a is the vertical acceleration. Solving for a, we get:

`a=(-F\sin35-mg)/(m)`

Then, replace F = 720 N, m = 68 kg, and g = 9.8 m/s² to get:

`\begin{gathered} a=\frac{-720N(\sin35)-(68\text{ kg\rparen\lparen9.8 m/s}^2}{68\text{ kg}} \\ \\ a=-15.9\text{ m/s}^2 \end{gathered}`

Therefore, the vertical acceleration is

A. -15.9 m/s²