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2014-10-12T14:42:01+05:30

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Let the required angle be β.

Range =  \frac{ v^{2}sin(2 \beta ) }{g}

maximum height =  \frac{ v^{2} sin^{2} \beta   }{2g}

So  \frac{ v^{2}sin(2 \beta ) }{g}  \frac{ v^{2} sin^{2} \beta }{2g}

 \frac{2 v^{2}sin \beta cos \beta  }{g}  =\frac{ v^{2}  sin^{2}  \beta }{2g}

⇒tan β = 4
⇒ β = arctan(4)
⇒ β = 75.96⁰

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2014-10-12T18:06:32+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
In kinematics we have the following formulas for the projectile motion:
It is useful to remember them or to derive them quickly.

time to reach the highest point = (u sin Ф /g)

R = 2 (u Sin Ф /g )(u cos Ф) = u² Sin 2Ф / g

H = (u sin Ф / g) (u Sin Ф/2) = u² Sin² Ф / 2g  

H / R = tan Ф / 4 

If H = R then    tan Ф = 4     =>  Ф = 75.96°  


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