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CAN YOU PLEASE SOLVE THE 13TH QUESTION AND I WILL RATE YOU.....
CAN YOU PLEASE SOLVE THE 13TH QUESTION AND I WILL RATE YOU.....


4 years ago

## Answers : (1)

Manas Shukla
102 Points
							U need to keep in mind that $v_{x}$ will be different for both the cases of 30 degrees and 60 degrees.Lets call them velocities $v_{x1}$ and $v_{x2}$ .For first case when angle = 30 degrees we can write$v_{x1}^{2} (tan^{2}45 - tan^{2}30) = 2gh$Since tan theta = y component / x componentSimilarly when angle of launch = 60 degrees,$v_{x2}^{2} (tan^{2}\Theta - tan^{2}60) = 2gh$Now keep in mind net velocity is common = vSo vcos(angle of launch) = Dividing both equations we get$v_{x2}^{2} (tan^{2}\Theta - tan^{2}60) = v_{x1}^{2} (tan^{2}45 - tan^{2}30)$$\cos ^{2}60 (tan^{2}\Theta - tan^{2}60) = cos^{2}30 (tan^{2}45 - tan^{2}30)$Simplifying we get $\Theta = tan^{-1} \sqrt{5}$

4 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