29 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.
Vy nikaal liya at half the height, then resultant. Last image ke baad kya karu.
Also resultant nikalna jaruri hai kya?
(B)?
I got 60° as the answer.
Yes
Just a sec. I'll share the soln.
Hint dedo after Vy kya karna hai
Soln mat bhejo
Aise kar sakte hai?
$$v{y{\frac{H}{2}}} = \sqrt{gH} = \frac{u sin\theta}{\sqrt{2}}
v{\frac{H}{2}} = \frac{u \sqrt{1+cos^2\theta}}{\sqrt{2}}
v{H}=u cos\theta
As per equation in question,
ucos\theta = \sqrt{\frac{2}{5}}\frac{u \sqrt{1+cos^2\theta}}{\sqrt{2}}$$
Opt
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Solve for cosθ
Is this right?
Nahi
V H/2 is resultant?
usinθ/√2 sirf y component hai H/2 par
Yup. Speed ke ratio hai na? Question me.
Yup
Then magnitude aka resultant is required
I see
Did you get it?
Stuck here
Write it as 1+cos²θ
Then square on both sides in the end
2cos²θ + sin²θ?
Yup
This is an identity?
cos²A+sin²A=1.....
Got it bro
Aagaya answer
+solved @Opt
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