Question

An airplane accelerates at 7.5 m/s2 at an angle of 16° above the horizontal. Find the horizontal and vertical components of the acceleration.
horizontal component: ______ m/s2
vertical component: ______ m/s2

Answer 1

Answer:

$the \: horizontal \: component \: of \: the \: acceleration : \\ a_{x} = 7.21\: {m(s)}^{ - 2} \\ the \: vertical \: component \: of \: the \: acceleration : \\ a_{y} = 2.07 \: {m(s)}^{ - 2}$

Explanation:

$the \: horizontal \: component \: of \: the \: acceleration : \\ a_{x} = a \: \cos( \alpha ) \\ a_{x} = 7.5 \: \cos(16) \\ a_{x} = 7.2094627195 \: {m(s)}^{ - 2} \\ \\ the \: vertical \: component \: of \: the \: acceleration : \\ a_{y} = a \: \sin( \alpha )\\a_{y} = 7.5 \: \sin(16) \\ a_{y} = 2.0672801686 \: {m(s)}^{ - 2}$

Answer 2

Answer:

horizontal component: $7.2\ m/s^2$

vertical component: $2.1\ m/s^2$

Explanation:

Rectangular components of a vector

Given a vector as a (magnitude, angle) pair, the rectangular components can be calculated as:

$a_x= magnitude*cos(angle)$

$a_y=magnitude*sin(angle)$

The acceleration of the airplane is given with a magnitude of 7.5 m/s^2 and an angle of 16°.

Calculate the components:

$a_x=7.5*cos(16^\circ)=7.2\ m/s^2$

$a_y=7.5*sin(16^\circ)=2.1\ m/s^2$

horizontal component: $7.2\ m/s^2$

vertical component: $2.1\ m/s^2$