Question

this question is from projectile motion


also answer my other questions​

Answer 1

Answer:

* here, at some place I have used "©" and 'alpha' because of unavailability of same symbols. But both means same.

the projectile is projected at an angle © with the horizontal with velocity u then the maximum height H attained by the projectile:

$H= \frac{ {u}^{2} { \sin( \alpha ) }^{2} }{2g} \\ \\ R = \frac{ {u}^{2} \sin(2 \alpha ) }{g} \\ so, \\ \frac{ {u}^{2} { \sin( \alpha ) }^{2} }{2g} = \frac{ {u}^{2} \sin(2 \alpha ) }{g} \\ \tan( \alpha ) = 4$

since, tan 76° = 4 [ given ]

so, © = 76°