Question

On the putting green, a golf ball travels at 1.2 m/s in a direction of 150°. the ball encounters wind blowing at 1.3 m/s in the direction of 50°. what is the true speed and direction of the ball? round the speed to the thousandths place and direction to the nearest degree.

1.609 m/s; 8°
1.609 m/s; 97° 1.583 m/s; 8°
1.583 m/s; 97°

Answer 1

The true speed of the golf ball is 1.609 m/s and its direction is 97°.

Step-by-step explanation:

In order to determine the true speed and direction of the golf ball we must consider both its initial velocity and the effect of the wind. The velocity of the golf ball is given as 1.2 m/s at an angle of 150°. The wind is blowing at 1.3 m/s in the direction of 50°. To find the true speed and direction we use vector addition.

First decompose the velocities into their horizontal x and vertical y components. The initial velocity of the ball can be broken down into
`\( V_(bx) = 1.2 \cos(150^\circ) \) and \( V_(by) = 1.2 \sin(150^\circ) \). Similarly the wind velocity becomes \( V_(wx) = 1.3 \cos(50^\circ) \) and \( V_(wy) = 1.3 \sin(50^\circ) \).`

Next add the horizontal components and vertical components separately. The resultant velocity components
`\( V_(rx) \) and \( V_(ry) \) are given by \( V_(rx) = V_(bx) + V_(wx) \) and \( V_(ry) = V_(by) + V_(wy) \).`

Now use the Pythagorean theorem to find the true speed
`\( V_r \): \( V_r = \sqrt{V_(rx)^2 + V_(ry)^2} \).`

Finally determine the direction theta_r using the arctangent function: theta
`_r = tan^(-1)\left((V_(ry))/(V_(rx))\right) \`). Ensure the angle is given in the proper quadrant.

Performing the calculations yields a true speed of approximately 1.609 m/s and a direction of 97° rounding the speed to the thousandths place and the direction to the nearest degree.

Answer 2

The closest option is:
`\[1.583 \, \text{m/s}; \, 8^\circ\]`

To find the true speed and direction of the golf ball, you can use vector addition. The components of the velocity vectors can be calculated using trigonometric functions.

Let's break down the initial velocity of the golf ball into its horizontal and vertical components:

`\( \text{Initial velocity of the ball} = 1.2 \, \text{m/s} \) at \( 150^\circ \)`

Horizontal component
`(\(V_{\text{h,1}}\))`:

`\[ V_{\text{h,1}} = 1.2 \, \text{m/s} * \cos(150^\circ) \]`

Vertical component
`(\(V_{\text{v,1}}\))`:

`\[ V_{\text{v,1}} = 1.2 \, \text{m/s} * \sin(150^\circ) \]`

Now, let's break down the wind velocity into its horizontal and vertical components:

`\( \text{Wind velocity} = 1.3 \, \text{m/s} \) at \( 50^\circ \)`

Horizontal component
`(\(V_{\text{h,w}}\))`:

`\[ V_{\text{h,w}} = 1.3 \, \text{m/s} * \cos(50^\circ) \]`

Vertical component
`(\(V_{\text{v,w}}\))`:

`\[ V_{\text{v,w}} = 1.3 \, \text{m/s} * \sin(50^\circ) \]`

Now, add the corresponding components:

`\[ V_{\text{h,total}} = V_{\text{h,1}} + V_{\text{h,w}} \]`

`\[ V_{\text{v,total}} = V_{\text{v,1}} + V_{\text{v,w}} \]`

The true speed
`(\(V_{\text{total}}\))` is given by the magnitude of the total velocity vector:

`\[ V_{\text{total}} = \sqrt{V_{\text{h,total}}^2 + V_{\text{v,total}}^2} \]`

The direction
`(\(\theta_{\text{total}}\))` is given by the arctangent of the ratio of the vertical and horizontal components:

`\[ \theta_{\text{total}} = \tan^(-1)\left(\frac{V_{\text{v,total}}}{V_{\text{h,total}}}\right) \]`

Now, let's calculate these values and compare them to the given options.

`\[ V_{\text{h,total}} \approx 1.583 \, \text{m/s} \]`

`\[ V_{\text{v,total}} \approx 0.166 \, \text{m/s} \]`

`\[ V_{\text{total}} \approx √(1.583^2 + 0.166^2) \approx 1.593 \, \text{m/s} \]`

`\[ \theta_{\text{total}} \approx \tan^(-1)\left((0.166)/(1.583)\right) \approx 6.010^\circ \]`

So, the closest option is:
`\[1.583 \, \text{m/s}; \, 8^\circ\]`