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        A ball thrown at different angles with the same speed u and from the same point and it has the same range in both cases. If y1 and y2 are the heights attained in two cases, then y1 + y2 is equal to	u^2/g	2u^2/g	u^2/2g	u^2/4g
3 years ago

Raman Mishra
67 Points
							Two projectiles have the same range if they are launched at angles which are complementary to each other. That is, if the 1st projectile is launched at an angle $\dpi{80} \theta$ from the horizontal, then the other projectile must be launched at an angle of $\dpi{80} \frac{\pi }{2} -\theta$ from the horizontal for having the same range.Now, maximum height of a projectile = $\dpi{80} h = \frac{u^{2}\sin ^{2}\theta}{2g}$ For the 1st ball we have $\dpi{80} h_{1} = \frac{u^{2}\sin ^{2}\theta}{2g}$For the 2nd ball, we have $\dpi{80} h_{2} = \frac{u^{2}\sin ^{2}(\frac{\pi }{2}-\theta )}{2g}$$\dpi{80} = \frac{u^{2}\cos ^{2}\theta }{2g}$       $\dpi{80} \therefore$   h1 + h2 = u2/2g

3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions