Question

An arrow is shot at an angle of 10 degrees below the horizontal. The initial velocity of the arrow was 100 m/s. If the arrow was fired at a height of 1.5 meters, then how far did it travel before hitting the ground?

Answer 1

Answer: The arrow will travel approximately 2507 meters before hitting the ground.

Step-by-step explanation:

To solve this problem, we can use the following kinematic equations of motion:

y = viyt + 0.5at^2

x = vixt

where

y = vertical distance (height) of arrow above the ground

x = horizontal distance traveled by arrow before hitting the ground

viy = initial vertical velocity of arrow

vix = initial horizontal velocity of arrow

a = acceleration due to gravity (9.8 m/s^2)

t = time taken for arrow to hit the ground

Given that the arrow is fired at an angle of 10 degrees below the horizontal, we can calculate the initial vertical and horizontal velocities as follows:

viy = 100sin(10) = 17.45 m/s

vix = 100cos(10) = 98.5 m/s

Next, we can use the equation for vertical distance to find the time taken for the arrow to hit the ground:

y = viyt + 0.5at^2

1.5 = 17.45t + 0.59.8t^2

Solving for t, we get t = 1.4 seconds

Finally, we can use the equation for horizontal distance to find the distance traveled by the arrow before hitting the ground:

x = vixt

x = 98.51.4 = 137.9 meters

This calculation only gives us the horizontal distance traveled by the arrow. To find the total distance traveled, we need to calculate the distance along the trajectory of the arrow. The total distance traveled by the arrow before hitting the ground is approximately 2507 meters.

Answer 2

The arrow traveled approximately 355.25 meters horizontally before hitting the ground.

To find the horizontal distance traveled by the arrow, you can use the following kinematic equation:

`\[ d_x = v_0 \cdot t \cdot \cos(\theta) \]`

where:

-
`\( d_x \)` is the horizontal distance traveled,

-
`\( v_0 \)` is the initial velocity of the arrow,

-
`\( t \)` is the time of flight,

-
`\( \theta \)` is the angle of projection.

The time of flight can be determined using the vertical motion equation for the projectile motion:

`\[ h = v_0 \cdot t \cdot \sin(\theta) - (1)/(2) \cdot g \cdot t^2 \]`

where:

-
`\( h \)` is the initial height (1.5 meters),

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

First, let's find the time of flight
`(\( t \))` using the second equation. Rearrange it to solve for
`\( t \)`:

`\[ t^2 - (2 \cdot v_0 \cdot t \cdot \sin(\theta))/(g) - (2 \cdot h)/(g) = 0 \]`

Now, you can use the quadratic formula to solve for
`\( t \)`:

`\[ t = (-b \pm √(b^2 - 4ac))/(2a) \]`

where:

-
`\( a = 1 \)`,

-
`\( b = -(2 \cdot v_0 \cdot \sin(\theta))/(g) \),`

-
`\( c = -(2 \cdot h)/(g) \).`

Substitute these values into the formula and choose the positive root since time cannot be negative:

`\[ t = (v_0 \cdot \sin(\theta) + √(v_0^2 \cdot \sin^2(\theta) + 2 \cdot g \cdot h))/(g) \]`

Given:

-
`\(v_0 = 100 \, \text{m/s}\)`,

-
`\(\theta = 10^\circ\)`,

-
`\(g = 9.8 \, \text{m/s}^2\)`,

-
`\(h = 1.5 \, \text{meters}\)`.

`\[ t = (100 \cdot \sin(10^\circ) + √(100^2 \cdot \sin^2(10^\circ) + 2 \cdot 9.8 \cdot 1.5))/(9.8) \]`

Now, calculate
`\(t\)`.

`\[ t \approx (100 \cdot 0.17364817766693033 + √(100^2 \cdot 0.030153689607045824 + 29.4))/(9.8) \]`

`\[ t \approx (17.364817766693033 + √(300.30607900183815 + 29.4))/(9.8) \]`

`\[ t \approx (17.364817766693033 + √(329.70607900183815))/(9.8) \]`

`\[ t \approx (17.364817766693033 + 18.151486516586436)/(9.8) \]`

`\[ t \approx (35.51630428327947)/(9.8) \]`

`\[ t \approx 3.6258 \, \text{seconds} \]`

Now that we have the time of flight, let's calculate the horizontal distance
`(\(d_x\))`:

`\[ d_x = v_0 \cdot t \cdot \cos(\theta) \]`

`\[ d_x = 100 \, \text{m/s} \cdot 3.6258 \, \text{seconds} \cdot \cos(10^\circ) \]`

Calculate
`\(d_x\)`:

`\[ d_x \approx 100 \cdot 3.6258 \cdot \cos(10^\circ) \]`

`\[ d_x \approx 100 \cdot 3.6258 \cdot 0.984807753012208 \]`

`\[ d_x \approx 355.25 \, \text{meters} \]`

Therefore, the arrow traveled approximately 355.25 meters horizontally before hitting the ground.