Question

Decent vehicle is traveling vertical@ 5.5 m/s and has horizontal velocity of 3.5 m/s . What speed and angle is descent path

Answer 1

Answer on Question #47792 – Physics - Mechanics | Kinematics | Dynamics

Decent vehicle is traveling vertical@ 5.5 m/s and has horizontal velocity of 3.5 m/s. What speed and angle is descent path

Solution:

$$\begin{array}{l} V_y = 5.5 \frac{m}{s} - \text{vertical component of the speed}; \\ V_x = 3.5 \frac{m}{s} - \text{horizontal component of the speed}; \\ V - \text{speed of the vehicle}; \\ \alpha - \text{angle between speed components}; \end{array}$$

Using Pythagoras's theorem for the right triangle, we can find speed of the vehicle:

$$V = \sqrt{V_x^2 + V_y^2} = \sqrt{\left(5.5 \frac{m}{s}\right)^2 + \left(3.5 \frac{m}{s}\right)^2} = 6.52 \frac{m}{s}$$

To find angle $\alpha$, we can use the tangent definition (from the right triangle):

$$\tan \alpha = \frac{V_y}{V_x}$$ $$\alpha = \arctan \left(\frac{V_y}{V_x}\right) = \arctan \left(\frac{5.5 \frac{m}{s}}{3.5 \frac{m}{s}}\right) = 57.5{}^\circ$$

Answer: speed: $6.52 \frac{m}{s}$, angle $57.5{}^\circ$;

http://www.AssignmentExpert.com/