Question

marble launcher shoots a marble horizontally from the height of 0.2 m above a horizontal floor. The marble lands on the floor 5 m away from the launcher. What is the initial speed of the marble?

Answer 1

Answer:

Approximately $25\; {\rm m\cdot s^{-1}}$ (assuming that air resistance on the marble is negligible, and that $g = 9.81\; {\rm m\cdot s^{-2}}$.)

Explanation:

Start by finding the duration $t$ of the flight of the marble.

If the air resistance on the marble is negligible, the marble will accelerate downward at a constant $a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}$.

Let $u_{y}$ denote the initial vertical velocity of the marble. Let $x_{y}$ denote the vertical displacement (change in height) of the marble.

$\displaystyle \frac{1}{2}\, a_{y}\, t^{2} + u_{y}\, t = x_{y}$.

Since the marble was launched horizontally, $u_{y} = 0\; {\rm m\cdot s^{-1}}$. This equation becomes:

$\displaystyle \frac{1}{2}\, a_{y}\, t^{2} = x_{y}$.

The marble was launched from a height of $0.2\; {\rm m}$ above the floor. When the marble lands, it would be $0.2\; {\rm m}\!$ below where it was launched. Hence, the vertical displacement of the flight will be $x_{y} = (-0.2)\; {\rm m}$.

Solve the equation for $t$:

$\begin{aligned} t^{2} = \frac{2\, x_{y}}{a_{y}}\end{aligned}$.

$\begin{aligned} t &= \sqrt{\frac{2\, x_{y}}{a_{y}}} \\ &= \sqrt{\frac{2\times (-0.2)\; {\rm m}}{(-9.81)\; {\rm m\cdot s^{-2}}}} \\ &\approx 0.202\; {\rm s}\end{aligned}$.

In other words, the marble was in the air for approximately $0.202\; {\rm s}$ before landing.

Also because the air resistance on the marble is negligible, the horizontal velocity of the marble will be constant during the entire flight.

The marble achieved a horizontal displacement $x_{x}$ of $5\; {\rm m}$ in that flight of approximately $0.202\; {\rm s}$. Hence, the (initial) horizontal velocity $u_{x}$ of the marble will be:

$\begin{aligned}u_{x} &= \frac{x_{x}}{t} \\ &= \frac{5\; {\rm m}}{0.202\; {\rm s}} \\ &\approx 25\; {\rm m\cdot s^{-1}}\end{aligned}$.

Combine both the initial horizontal and vertical velocity of the marble to find the initial speed $u$:

$\begin{aligned} u &= \sqrt{{u_{x}}^{2} + {u_{y}}^{2}} \\ &\approx \sqrt{(25\; {\rm m\cdot s^{-1}})^{2} + (0\; {\rm m\cdot s^{-1}})^{2}} \\ &= 25\; {\rm m\cdot s^{-1}}\end{aligned}$.