the vertical and horizontal components of Earth's magnetic induction at a place are 2 into 10 raise to minus 5 and 3.464 into 10 raise to minus 5 respectively calculate angle of dip at the place
Answers
Explanation:
We will divide the Projectile motion into 2 simultaneously occuring 1D motions.
Along X axis :
\sf{ \therefore \: x = u \cos( \theta) \times t}∴x=ucos(θ)×t
\sf{ \implies \: x = 20 \cos( 60 \degree) \times 1}⟹x=20cos(60°)×1
\sf{ \implies \: x = 20 \times \dfrac{1}{2} \times 1}⟹x=20×
2
1
×1
\sf{ \implies \: x = 10 \: m}⟹x=10m
Along Y axis :
\sf{ \therefore \: y = u \sin( \theta) t - \frac{1}{2} g {t}^{2}}∴y=usin(θ)t−
2
1
gt
2
\sf{ \implies \: y = 20 \sin( 60 \degree) \times 1- ( \frac{1}{2} \times 10 \times {1}^{2}} )⟹y=20sin(60°)×1−(
2
1
×10×1
2
)
\sf{ \implies \: y = 20 \times \frac{ \sqrt{3} }{2} - 5}⟹y=20×
2
3
−5
\sf{ \implies \: y =17.32 - 5}⟹y=17.32−5
\sf{ \implies \: y =12.32 \: m}⟹y=12.32m
So , coordinate will be :
\boxed{ \red{ \bold{ \sf{ \huge{(x,y) = (10,12.32)}}}}}
(x,y)=(10,12.32)