Grade 11General Physics

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

MANOJ SAINI

9 Years agoGrade 11

1 Answer

hridya

9 Years ago

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

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

MANOJ SAINI

1 Answer

hridya

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