Answer
(h) The $y$ component of the ant’s net displacement is $0.802$ $m$
Work Step by Step
According to the given data,
$\vec d_{1}=d_{1}\cos45^{\circ}(-\hat i)+d_{1}\sin45^{\circ}\hat j=-0.40\cos45^{\circ}\hat i+0.40\sin45^{\circ}\hat j$
$\vec d_{2}=d_{2}\cos0^{\circ}\hat i+d_{2}\sin0^{\circ}\hat j=0.50\hat i$
$\vec d_{3}=d_{3}\cos60^{\circ}\hat i+d_{3}\sin60^{\circ}\hat j=0.60\cos60^{\circ}\hat i+0.60\sin60^{\circ}\hat j$
Therefore, the resultant displacement vecotor takes the form
$\vec d=\vec d_{1}+\vec d_{2}+\vec d_{3}$
or, $\vec d=(-0.40\cos45^{\circ}\hat i+0.40\sin45^{\circ}\hat j)+0.50\hat i+(0.60\cos60^{\circ}\hat i+0.60\sin60^{\circ}\hat j)$
or, $\vec d=(-0.40\cos45^{\circ}+0.50+0.60\cos60^{\circ})\hat i+(0.40\sin45^{\circ}+0.60\sin60^{\circ})\hat j$
or, $\vec d=0.517\hat i+0.802\hat j$
Thus, the $y$ component of the ant’s net displacement is $0.802$ $m$
