Question

1. in the figure, a small, solid, uniform ball is to be shot from point p so that it rolls smoothly along a horizontal path, up along a ramp, and onto a plateau. then it leaves the plateau horizontally to land on a game board, at a horizontal distance d from the right edge of the plateau. the vertical heights are h₁ = 5.00 cm and h₂ = 1.60 cm. with what speed must the ball be shot at point p for it to land at d = 6.00 cm? show all work.

Explanation:

Step1: Analyze the projectile motion from the plateau

The ball leaves the plateau horizontally, so the vertical motion is free - fall with initial vertical velocity \(v_{y0} = 0\). The vertical displacement is \(h_2\), and we use the equation \(h_2=\frac{1}{2}gt^{2}\) to find the time of flight \(t\).
Given \(h_2 = 1.60\space cm=0.016\space m\) and \(g = 9.8\space m/s^{2}\).
From \(h_2=\frac{1}{2}gt^{2}\), we can solve for \(t\):

$$t=\sqrt{\frac{2h_2}{g}}=\sqrt{\frac{2\times0.016}{9.8}}\approx\sqrt{\frac{0.032}{9.8}}\approx\sqrt{0.003265}\approx0.0571\space s$$

The horizontal motion is uniform motion with \(d = v_x t\), where \(v_x\) is the horizontal velocity when the ball leaves the plateau. Given \(d = 6.00\space cm = 0.06\space m\), we can find \(v_x\):

$$v_x=\frac{d}{t}=\frac{0.06}{0.0571}\approx1.05\space m/s$$

Step2: Use conservation of mechanical energy from point \(P\) to the plateau

The ball is a solid uniform ball, so its moment of inertia \(I=\frac{2}{5}mr^{2}\). The total mechanical energy at point \(P\) is \(E_{P}=\frac{1}{2}mv_{P}^{2}+\frac{1}{2}I\omega_{P}^{2}\) (since it is rolling without slipping, \(\omega_{P}=\frac{v_{P}}{r}\)). At the plateau, the total mechanical energy is \(E_{plateau}=mgh_1+\frac{1}{2}mv_{x}^{2}+\frac{1}{2}I\omega_{x}^{2}\) (and \(\omega_{x}=\frac{v_{x}}{r}\) because of no - slipping).
Since mechanical energy is conserved (\(E_{P}=E_{plateau}\)):

$$\frac{1}{2}mv_{P}^{2}+\frac{1}{2}\times\frac{2}{5}mr^{2}\times(\frac{v_{P}}{r})^{2}=mgh_1+\frac{1}{2}mv_{x}^{2}+\frac{1}{2}\times\frac{2}{5}mr^{2}\times(\frac{v_{x}}{r})^{2}$$

Simplify the equation. The \(m\) and \(r\) terms cancel out:

$$\frac{1}{2}v_{P}^{2}+\frac{1}{5}v_{P}^{2}=gh_1+\frac{1}{2}v_{x}^{2}+\frac{1}{5}v_{x}^{2}$$

$$\frac{5 + 2}{10}v_{P}^{2}=gh_1+\frac{5+2}{10}v_{x}^{2}$$

$$\frac{7}{10}v_{P}^{2}=gh_1+\frac{7}{10}v_{x}^{2}$$

Multiply both sides by \(\frac{10}{7}\):

$$v_{P}^{2}=\frac{10}{7}gh_1 + v_{x}^{2}$$

Given \(h_1=5.00\space cm = 0.05\space m\), \(g = 9.8\space m/s^{2}\) and \(v_x\approx1.05\space m/s\)
First, calculate \(\frac{10}{7}gh_1\):

$$\frac{10}{7}\times9.8\times0.05=\frac{10\times9.8\times0.05}{7}=\frac{4.9}{7} = 0.7$$

Then, \(v_{x}^{2}=(1.05)^{2}\approx1.1025\)

$$v_{P}^{2}=0.7 + 1.1025=1.8025$$

$$v_{P}=\sqrt{1.8025}\approx1.34\space m/s$$

Answer:

The speed at which the ball must be shot at point \(P\) is approximately \(\boldsymbol{1.34\space m/s}\)