A boat heading north crosses a wide river with speed of 10kmhr relative to water the has 5kmhr due east what is velocity w respect stationary ground observer?

The velocity of the boat with respect to the stationary ground observer can be found using vector addition. We can represent the velocity of the boat relative to the water as a vector \(\overrightarrow{v_b}\) of magnitude 10 km/hr at an angle of 0° (since the boat is heading north). The velocity of the water relative to the ground can be represented as a vector \(\overrightarrow{v_w}\) of magnitude 5 km/hr at an angle of 90° (since the water is flowing east).

To find the velocity of the boat with respect to the ground, we add the two vectors:

$$\overrightarrow{v} = \overrightarrow{v_b} + \overrightarrow{v_w}$$

Using the law of cosines, we can find the magnitude of the resultant vector:

$$v = \sqrt{v_b^2 + v_w^2 + 2v_bv_w\cos\theta}$$

where \(\theta\) is the angle between the two vectors. Substituting the given values, we get:

$$v = \sqrt{10^2 + 5^2 + 2(10)(5)\cos90°}$$

$$v = \sqrt{100 + 25 + 0}$$

$$v = \sqrt{125}$$

$$v = 11.18 \text{ km/hr}$$

To find the angle of the resultant vector, we can use the law of sines:

$$\sin\theta = \frac{v_w\sin\theta}{v}$$

Substituting the given values, we get:

$$\sin\theta = \frac{5\sin90°}{11.18}$$

$$\sin\theta = \frac{5}{11.18}$$

$$\theta = \sin^{-1}\left(\frac{5}{11.18}\right)$$

$$\theta = 26.57°$$

Therefore, the velocity of the boat with respect to the stationary ground observer is 11.18 km/hr at an angle of 26.57° north of east.