91Ó°ÊÓ

Kinematic equations are the essential tools you need when dealing with projectile motion problems, like the one with the girl on the rope swing. These equations describe the motion of objects and help us find unknown variables such as position, velocity, or time. In this exercise, the primary kinematic equation we used was related to vertical motion: \[ y = v_{y0} \cdot t + \frac{1}{2}a t^2 \]This equation helps us find the vertical position of the object, considering its initial vertical velocity, time in the air, and acceleration due to gravity. Let’s break it down:

• \(y\) is the final vertical position. In our scenario, it's 0 because the girl lands at the water's surface.
• \(v_{y0}\) is the initial vertical velocity obtained by breaking down the initial velocity into components.
• \(t\) is the time the girl is in motion, given as 0.616 seconds.
• \(a\) is the vertical acceleration, which is -9.81 m/s² due to gravity.
• \(y_0\) is the initial height we're solving for.

By rearranging the equation, we can isolate \(y_0\) and find how high above the water she was initially.

When dealing with projectile motion, such as a swing above water, it's essential to understand how to break down initial velocity into components. The velocity given in the problem typically needs to be separated into horizontal and vertical parts to make accurate calculations.For the girl swinging above the swimming hole, her initial velocity is 2.25 m/s at a 35° angle above the horizontal. We use trigonometric functions to split this velocity into:

• Horizontal component: \( v_{x0} = v_0 \cos(\theta) \)
• Vertical component: \( v_{y0} = v_0 \sin(\theta) \)

Calculating Each Component

• For the horizontal component: \(v_{x0} = 2.25 \cos(35°) \approx 1.84\, \mathrm{m/s}\)
• For the vertical component: \(v_{y0} = 2.25 \sin(35°) \approx 1.29\, \mathrm{m/s}\)

These components help us use the kinematic equations effectively. Specifically, the vertical component is used in the equation for the motion to find how high the girl was when she let go.

Vertical motion is a crucial aspect in projectile problems, as it tells us how high or low an object travels over time. In this exercise, the girl’s motion becomes a vertical motion problem once she lets go of the rope.The vertical motion is solely influenced by gravity, which accelerates the girl downward at \(-9.81\,\mathrm{m/s}^2\). Understanding this helps us predict the motion's outcome and set up the necessary equations.

Equation in Action

Using the vertical motion equation \[ y = v_{y0} \cdot t + \frac{1}{2} a t^2 \]we accounted the initial vertical velocity (1.29 m/s) she had from the swing. Over the time in the air (0.616 seconds), the impact of gravity needed calculation. Solving this equation allows us to discover the initial height, thanks to her known descent back to zero level at the water's surface.Recognizing gravity's effect and calculating times correctly is priceless in these problems, helping determine the initial conditions we often seek in projectile trajectory tasks.