Question

question 4
6 points
wanting to measure how fast you can hit a cue ball when you break. so you set up table, 0.93 m above the floor, strike the cue ball, and it flies a horizontal distance of 3.4 m before hitting the floor. with what speed can you hit a cue ball?

Explanation:

Step1: Analyze vertical - motion

The ball is in free - fall in the vertical direction. The initial vertical velocity $v_{0y}=0$ m/s, the acceleration $a = g=9.8$ m/s², and the vertical displacement $y - y_0=- 0.93$ m. Using the equation $y - y_0=v_{0y}t+\frac{1}{2}at^{2}$, since $v_{0y} = 0$, we have $y - y_0=\frac{1}{2}at^{2}$.
$-0.93=\frac{1}{2}\times(-9.8)t^{2}$

Step2: Solve for time $t$

$t^{2}=\frac{2\times0.93}{9.8}$
$t=\sqrt{\frac{2\times0.93}{9.8}}\approx\sqrt{\frac{1.86}{9.8}}\approx0.435$ s

Step3: Analyze horizontal - motion

In the horizontal direction, there is no acceleration ($a_x = 0$), and the horizontal displacement $x - x_0 = 3.4$ m. The horizontal velocity $v_x$ is constant, and we use the equation $x - x_0=v_x t$. Then $v_x=\frac{x - x_0}{t}$.
$v_x=\frac{3.4}{0.435}\approx7.82$ m/s

Answer:

7.82 m/s