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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?

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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)}.)

Step-by-step 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 (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 (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) = (2\, x_(y))/(a_(y))\end{aligned}.


\begin{aligned} t &= \sqrt{(2\, x_(y))/(a_(y))} \\ &= \sqrt{\frac{2* (-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) &= (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}.

User Bryan Bende
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