Answer:22 m/s
Explanation:
Given
launch angle [tex]\theta =60^{\circ}[/tex]
height of goal [tex]h=4.55\ m[/tex]
and horizontal distance [tex]x=40\ m[/tex]
Suppose initial speed is [tex]u[/tex]
Trajectory of a Projectile is given by
[tex]y=x\tan \theta -\frac{1}{2}\frac{gx^2}{u^2\cos ^2\theta }[/tex]
substituting the values we get
[tex]4.55=40\tan (60)-0.5\times \frac{9.8\times (40)^2}{u^2\cdot \cos ^260 }[/tex]
[tex]4.55=69.28-0.5\times \frac{15,680}{u^2\cdot 0.25}[/tex]
[tex]\frac{31,360}{u^2}=69.28-4.55[/tex]
[tex]\frac{31,360}{64.73}=u^2[/tex]
[tex]u^2=484.47[/tex]
[tex]u=22.01\ m/s[/tex]
So, initial launch speed is [tex]22\ m/s[/tex]
