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Grade: 11

                        

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 component
Similarly when angle of launch = 60 degrees,
v_{x2}^{2} (tan^{2}\Theta - tan^{2}60) = 2gh
Now keep in mind net velocity is common = v
So 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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