Question

1. the landing speed of the space shuttle columbia is 347 km/hr. if the shuttle is landing at an angle of 15.0° with respect to the horizontal, what are the horizontal and vertical components of its velocity?

Explanation:

Step1: Recall component - formula

Let the speed of the shuttle be $v = 347$ km/hr and the angle with the horizontal be $\theta=15.0^{\circ}$. The horizontal component of the velocity $v_x$ is given by $v_x = v\cos\theta$ and the vertical component $v_y$ is given by $v_y = v\sin\theta$.

Step2: Calculate horizontal component

$v_x=v\cos\theta = 347\times\cos(15.0^{\circ})$. Since $\cos(15.0^{\circ})=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659$, then $v_x = 347\times0.9659\approx335.2$ km/hr.

Step3: Calculate vertical component

$v_y = v\sin\theta=347\times\sin(15.0^{\circ})$. Since $\sin(15.0^{\circ})=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.2588$, then $v_y = 347\times0.2588\approx89.9$ km/hr.

Answer:

Horizontal component: approximately $335.2$ km/hr, Vertical component: approximately $89.9$ km/hr