The initial velocity of the ball is 8.2 m/s
Explanation:
The motion of the ball is a projectile motion, which consists of two separate motions:
- A uniform motion along the horizontal direction, at constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration ([tex]g=9.8 m/s^2[/tex], acceleration of gravity)
The range of a projectile can be derived by the equation of motions along the two directions, and it is found to be:
[tex]d=\frac{v^2 sin 2\theta}{g}[/tex]
where
v is the initial velocity of the projectile
[tex]\theta[/tex] is the angle of projection
g is the acceleration of gravity
For the ball in this problem, we have
[tex]\theta=30^{\circ}[/tex]
d = 6 m is the range
Solving for v, we find the initial velocity:
[tex]v=\sqrt{\frac{gd}{sin 2\theta}}=\sqrt{\frac{(9.8)(6)}{sin (2\cdot 30^{\circ})}}=8.2 m/s[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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