Question

Part A A helicopter is flying horizontally with a speed of 909 m/s over a hill that slopes upward with a 2% grade (that is, the "rise" is 2% of the "run"). What is the component of the helicopter's velocity perpendicular to the sloping surface of the hill? O 18 m/s O 450 m/s 909 m/s O 818 m/s

PART B)

Which one is true answer? Thanks all.

Answer 1

Answer : --

(a)

2 tano 100

$sin\theta = \frac{2} { \sqrt{2^{2}+100^{2}}}$

$sin\theta = \frac{2} {100.0199}$

so the verticle componant of the velocity = $vsin\theta = 909 \times\frac{2} {100.0199}$

$vsin\theta = 18.17$ m/s

so the option (a)is correct .

(b)$cos \theta = \frac{\vec{A} \cdot \vec{ B}}{ AB}$

• $cos \theta = \frac{2\times -2 + 4 \times 2}{ \sqrt{8}\sqrt{20}}$
• $cos \theta = 0.316227445$
• $\theta = 71.5651$
• option(D ) is correct .