Question

plz guys help .....
its for tomorrow's board exam
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Answer 1

Answer:

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)