Respuesta :
Answer:
Explanation:
Given
zero vertical displacement
suppose \theta is the inclination angle and u is the initial velocity
considering horizontal motion
[tex]u_x=u\cos \theta [/tex]
range is given by
[tex]R=u\cos \theta \times t[/tex]
where t is the time of flight
Now consider vertical motion
[tex]u_y=u\sin \theta [/tex]
[tex]y=ut+\frac{1}{2}at^2[/tex]
a=acceleration
for zero displacement y=0
[tex]0=u\sin \theta \times t-\frac{1}{2}gt^2[/tex]
[tex]t=\frac{2u\sin \theta }{g}[/tex]
substitute the value of t in Range
[tex]R=u\cos \theta \times \frac{2u\sin \theta }{g}[/tex]
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
for maximum vale of [tex]R \sin 2\theta [/tex] must be equal to 1
i.e. [tex]\sin 2\theta =1[/tex]
[tex]2\theta =\frac{\pi }{2}[/tex]
[tex]\theta =\frac{\pi }{4}[/tex]
