Answer:
Explanation:
Given
launching inclination is [tex]\alpha[/tex]
suppose u is the initial velocity
vertical velocity [tex]v_y=u\sin \alpha[/tex]
horizontal velocity [tex]v_x=u\cos \alpha[/tex]
Vertical distance traveled by Projectile is
[tex]y=v_y\times t-\frac{1}{2}gt^2----1[/tex]
Now equation of line passing through origin
[tex]y=x\tan \beta------2[/tex]
Equating two equation we get
[tex]x\tan \beta=v_y\times t-\frac{1}{2}gt^2[/tex]
putting [tex]x=v_x\times t=u\cos \alpha \times t[/tex]
[tex]u\cos \alpha \tan \beta \cdot t=v_y\times t-\frac{1}{2}gt^2[/tex]
[tex]t\left [ u\sin \alpha -u\cos \alpha \tan\beta -\frac{gt}{2}\right ]=0[/tex]
[tex]t=\frac{2u\left [ \sin \alpha -\cos \alpha \tan \beta \right ]}{g}[/tex]
