Answer :

TPS
Let the required angle be β.

Range = [tex] \frac{ v^{2}sin(2 \beta ) }{g} [/tex]

maximum height = [tex] \frac{ v^{2} sin^{2} \beta }{2g} [/tex]

So [tex] \frac{ v^{2}sin(2 \beta ) }{g} [/tex] = [tex] \frac{ v^{2} sin^{2} \beta }{2g} [/tex]

⇒[tex] \frac{2 v^{2}sin \beta cos \beta }{g} =\frac{ v^{2} sin^{2} \beta }{2g} [/tex]

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

kvnmurty
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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