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Grade: 11

                        

the range and height of a projectile are 90m and 45m respectively. the angle of projection and projection velocity is???

4 years ago

Answers : (1)

hridya
38 Points
							
range =R={\frac  {v^{2}\sin 2\theta }{g}}=90----------(1”)
height=h={\frac  {v_{0}^{2}\sin ^{2}(\theta )}{2g}}=45-------------(2)multiply (2) with 2
h={\frac  {v_{0}^{2}\sin ^{2}(\theta )}{2g}}*2=90-----------------(3)
equate 1 and 3 u will get value of "θ" which angle of projection from it find value of sin"θ" and put in equation2 u will get the velocity of projection..hope the answer was helpful to you
 
 
4 years ago
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