Question

A potato gun is fired horizontally from a height of 1.50 meters with the potato launched at 25.0 m/s.

How far from the gun will the potato

Answer 1

The potato, launched horizontally at 25.0 m/s from a 1.50-meter height, will travel approximately 13.825 meters.

To determine the horizontal distance (dx) traveled by the potato after being fired from the potato gun, we can use the kinematic equation:

`\[ dx = v_(0x) \cdot t \]`

where:

- dx is the horizontal distance,

-
`\( v_(0x) \)` is the horizontal component of the initial velocity (which is the launch velocity, 25.0 m/s),

- t is the time of flight.

The time of flight (t) can be found using the vertical motion equation for free fall:

`\[ h = v_(0y) \cdot t + (1)/(2) \cdot g \cdot t^2 \]`

where:

- h is the initial height (1.50 meters),

-
`\( v_(0y) \)` is the vertical component of the initial velocity (0 m/s for horizontal launch),

- g is the acceleration due to gravity (approximately 9.8 m/s²).

Rearranging the equation for time (t), we get:

`\[ t = \sqrt{(2h)/(g)} \]`

Now, substitute the given values:

`\[ t = \sqrt{\frac{2 \cdot 1.50 \, \text{m}}{9.8 \, \text{m/s}^2}} \]`

Calculate t and then use it to find dx using the first kinematic equation.

Let's calculate t first:

`\[ t = \sqrt{\frac{2 \cdot 1.50 \, \text{m}}{9.8 \, \text{m/s}^2}} \]`

`\[ t \approx \sqrt{(3.00)/(9.8)} \]`

`\[ t \approx \sqrt{0.306 \, \text{s}^2} \]`

`\[ t \approx 0.553 \, \text{s} \]`

Now, use this time value to find dx:

`\[ dx = v_(0x) \cdot t \]`

`\[ dx = 25.0 \, \text{m/s} \cdot 0.553 \, \text{s} \]`

`\[ dx \approx 13.825 \, \text{m} \]`

So, the potato will travel approximately 13.825 meters horizontally from the gun.