Question

1. a ball is rolling off a table.

a. if it takes 0.25 seconds for the ball to strike the ground, how high is the table?
b. if the ball lands 1.5 meters from the table, how fast was the ball rolling when it fell off the table?
c. what is the final y velocity of the ball right before it hits the ground?
d. what is the magnitude of the balls resultant velocity when it hits the ground?

Explanation:

Step1: Calculate the height of the table (a)

Use the equation \(h = v_{0y}t+\frac{1}{2}gt^{2}\). Since \(v_{0y} = 0\) (initial vertical velocity is zero for a ball rolling off a table), \(h=\frac{1}{2}gt^{2}\). Given \(t = 0.25\ s\) and \(g=9.8\ m/s^{2}\), then \(h=\frac{1}{2}\times9.8\times(0.25)^{2}\).

$$h = 0.30625\ m$$

Step2: Calculate the initial horizontal velocity (b)

In the horizontal direction, \(x = v_{0x}t\) (no acceleration \(a_x = 0\)). Given \(x = 1.5\ m\) and \(t=0.25\ s\), then \(v_{0x}=\frac{x}{t}\).

$$v_{0x}=\frac{1.5}{0.25}=6\ m/s$$

Step3: Calculate the final vertical velocity (c)

Use the equation \(v_y=v_{0y}+gt\). Since \(v_{0y} = 0\), \(v_y = gt\). Given \(t = 0.25\ s\) and \(g = 9.8\ m/s^{2}\), then \(v_y=9.8\times0.25\).

$$v_y = 2.45\ m/s$$

Step4: Calculate the resultant velocity (d)

The resultant velocity \(v=\sqrt{v_{x}^{2}+v_{y}^{2}}\). We know \(v_{x}=v_{0x} = 6\ m/s\) (horizontal velocity is constant) and \(v_{y}=2.45\ m/s\). Then \(v=\sqrt{6^{2}+2.45^{2}}\).

$$v=\sqrt{36 + 6.0025}=\sqrt{42.0025}\approx6.48\ m/s$$

Answer:

a. The height of the table is \(0.30625\ m\).
b. The ball was rolling at \(6\ m/s\) when it fell off the table.
c. The final \(y -\) velocity of the ball right before it hits the ground is \(2.45\ m/s\).
d. The magnitude of the ball’s resultant velocity when it hits the ground is approximately \(6.48\ m/s\).