Answer:
Explanation:
Given
Speed of ball [tex]u=3.2\ m/s[/tex]
Plane is inclined at an angle [tex]20^{\circ}[/tex]
To win the Game we need to hit the target at [tex]x=2.4\ m[/tex] away
Launch angle of ball [tex]\theta [/tex]
Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane
So Net acceleration in vertical plane is [tex]g\sin 20[/tex]
Range of Projectile is given by
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
for [tex]R=2.4\ m[/tex]
[tex]2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}[/tex]
[tex]\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}[/tex]
[tex]\sin 2\theta =0.7855[/tex]
[tex]2\theta =51.77[/tex]
[tex]\theta =25.88^{\circ}[/tex]
so ball must be launched at an angle of [tex]25.88^{\circ}[/tex]
