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A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4.8 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a 25 g plastic ball. The two balls collide 1.0 m above the floor. Part A What was the launch speed of the plastic ball?

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Answer

given,

mass of the steel ball = 35 g = 0.035 kg

the horizontal distance is

( v cos θ ) t = 4.8..........(1)

the vertical distance is

[tex](v sin \theta ) t - \dfrac{1}{2}gt^2 = 1.2[/tex]...(2)

since,

[tex](v sin \theta )t = 4 [/tex]

[tex]\dfrac{1}{2}gt^2 = (v sin \theta) t - 1.2[/tex]

[tex]\dfrac{1}{2}gt^2 = 4 - 1.2[/tex]..................(3)

using equation(2) and (3)

(v t)² (cos²θ+ sin²θ) = 4.8² + 4²

(v t)²  = 39.04

[tex]\dfrac{1}{2}gt^2 = 4 - 1.2[/tex]

[tex]t = \sqrt{\dfrac{2(4-1.2)}{9.8}}[/tex]

t = 0.76 s

the speed is

vt = 6.25

v × 0.76 = 6.25

v = 8.22 m/s

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