Explanation:
Initial Velocity (u) = 11 m/sec
Angle of Projectile (theta) = 15.3°
(Ques - 1) How far does he jump?
(Ans - 1) Find Horizontal Range :
[tex]R = \frac{ {u}^{2} \sin(2 \alpha ) }{g} = \frac{(11)^{2} \sin(30.6°) }{9.8} [/tex]
[tex]Range = 6.17 \: m[/tex]
(Ques - 2) What maximum height does he reach?
(Ans - 2)
[tex]H = \frac{ {u}^{2} {sin }^{2} \alpha }{2g} = \frac{(11)^{2} {sin}^{2}15.3° }{19.6} [/tex]
[tex]Maximum \: Height = 0.41 \: m[/tex]
