Answer:
6.0621 m/s
Explanation:
we all know that horizontal component of the velocity in projectile motion is
= u cosα
u= initial speed of 7.00 m/s of the gun.
α = angle of the initial velocity with the horizontal = 30°
therefore, horizontal component of the ball's initial velocity = 7×cos30°
[tex]7\times\frac{\sqrt{3} }{2} \\=6.0621 \text{m/s}[/tex]
