Question

multiple choice 1 point
a student is attempting to hit a target 4 m from the edge of a desk that is 1.4 m tall. the marble rolls off a table horizontally as shown below.
according to the above information and diagram, what velocity must the marble have to strike the target?
7.49 m/s
1.55 m/s
14.0 m/s
2.86 m/s

Explanation:

Step1: Analyze vertical - motion

The vertical - motion of the marble is a free - fall motion. The height of the table \(h = 1.4\ m\), and the initial vertical velocity \(v_{0y}=0\ m/s\). Using the equation \(h = v_{0y}t+\frac{1}{2}gt^{2}\), since \(v_{0y} = 0\ m/s\), we have \(h=\frac{1}{2}gt^{2}\).

Step2: Solve for time \(t\)

From \(h=\frac{1}{2}gt^{2}\), we can solve for \(t\): \(t=\sqrt{\frac{2h}{g}}\). Given \(h = 1.4\ m\) and \(g = 9.8\ m/s^{2}\), then \(t=\sqrt{\frac{2\times1.4}{9.8}}\approx\sqrt{\frac{2.8}{9.8}}\approx\sqrt{\frac{2}{7}}\approx0.53\ s\).

Step3: Analyze horizontal - motion

The horizontal distance \(x = 4\ m\). In horizontal motion (no acceleration, \(a_x = 0\)), the equation is \(x=v_{0x}t\). We know \(x\) and \(t\), and we want to find \(v_{0x}\). So \(v_{0x}=\frac{x}{t}\).

Step4: Calculate initial horizontal velocity

Substitute \(x = 4\ m\) and \(t\approx0.53\ s\) into \(v_{0x}=\frac{x}{t}\), we get \(v_{0x}=\frac{4}{0.53}\approx7.55\ m/s\approx7.49\ m/s\) (due to rounding differences).

Answer:

7.49 m/s