Question

instead of dropping the mic, you let it roll off the table with an initial velocity of 0.5m/s. if the table is 0.76 m high, a. how long did it take to reach the ground? b. what is the mics range?

Explanation:

Step1: Analyze vertical - motion

The vertical - motion of the mic is a free - fall motion. The initial vertical velocity $v_{0y}=0\ m/s$, the acceleration due to gravity $g = 9.8\ m/s^{2}$, and the vertical displacement $y=- 0.76\ m$ (taking downwards as negative). We use the equation $y = v_{0y}t+\frac{1}{2}at^{2}$. Since $v_{0y} = 0$, the equation simplifies to $y=\frac{1}{2}at^{2}$, where $a=-g$.

Step2: Solve for time $t$

From $y=\frac{1}{2}at^{2}$, we can solve for $t$. Rearranging the formula gives $t=\sqrt{\frac{-2y}{g}}$. Substituting $y = - 0.76\ m$ and $g = 9.8\ m/s^{2}$ into the formula:
\[t=\sqrt{\frac{-2\times(-0.76)}{9.8}}=\sqrt{\frac{1.52}{9.8}}\approx0.39\ s\]

Step3: Analyze horizontal - motion

The horizontal motion of the mic is a uniform - motion with constant velocity because there is no horizontal acceleration ($a_x = 0$). The initial horizontal velocity $v_{0x}=0.5\ m/s$. The horizontal displacement (range) $x$ is given by the formula $x = v_{0x}t$.

Step4: Calculate the range $x$

We know $v_{0x}=0.5\ m/s$ and $t\approx0.39\ s$. Substituting these values into the formula $x = v_{0x}t$, we get $x=0.5\times0.39 = 0.195\ m$.

Answer:

a. $t\approx0.39\ s$
b. $x = 0.195\ m$