The maximum height of the ball is 1.45 m
While in the barrel, the two forces acting on the ball are its weight W = mg and the force due to compressed air, F.
While in the barrel, the net force acting on the ball is F' = F - mg.
Since the compressed air exerts a force F = 3.2 N on the ball and its mass, m = 56 g = 0.056 kg and g = acceleration due to gravity = 9.8 m/s².
So, F' = F - mg
= 3.2 N - 0.056 kg × 9.8 m/s²
= 3.2 N - 0.5488 N
= 2.6512 N.
We now find the acceleration of the ball in the barrel, a from F' = ma
a = F'/m
= 2.6512 N/0.056 kg
= 47.34 m/s²
We now need to find the velocity with which the ball leaves the barrel from v² = u² + 2as where u = initial velocity of ball = 0 m/s, a = acceleration of ball = 47.34 m/s² and s = length of barrel = 30 cm = 0.30 m
So, substituting the values of the variables into the equation, we have
v² = u² + 2as
v² = (0 m/s)² + 2 × 47.34 m/s² × 0.30 m
v² = 0 m²/s² + 28.404 m²/s²
v² = 28.404 m²/s²
v = √28.404 m²/s²
v = 5.33 m/s
Finally, we find the maximum height the ball reaches after leaving the barrel from
v'² = v² - 2gh where v = initial velocity of ball after leaving the barrel = 5.33 m/s, v' = final velocity of ball at maximum height = 0 m/s (since it is momentarily at rest at maximum height),g = acceleration of ball = acceleration due to gravity = 9.8 m/s² and h = maximum height of ball.
Since we require h, we make h subject of the formula.
So, h = -(v'² - v²)/2g
Substituting the values of the variables into the equation, we have
h = -(v'² - v²)/2g
h = -((0 m/s)² - (5.33 m/s)²)/2(9.8 m/s²)
h = -((0 m²/s² - 28.41 m²/s²)/19.6m/s²)
h = -(- 28.41 m²/s²)/19.6m/s²)
h = 28.41 m²/s²)/19.6m/s²)
h = 1.45 m
So, the maximum height of the ball is 1.45 m
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