Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The distance traveled in horizontal direction is [tex]D = 1.38 m[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]L = 1.6 \ m[/tex]
The mass of the ball is [tex]m = 200 g = \frac{200}{1000} = 0.2 \ kg[/tex]
The height of ball is [tex]h = 1.5 \ m[/tex]
Generally the work energy theorem can be mathematically represented as
[tex]PE = KE[/tex]
Where PE is the loss in potential energy which is mathematically represented as
[tex]PE =mgh[/tex]
Where h is the difference height of ball at A and at B which is mathematically represented as
[tex]h = y_A - y_B[/tex]
So [tex]PE =mg(y_A - y_B)[/tex]
While KE is the gain in kinetic energy which is mathematically represented as
[tex]KE = \frac{1}{2 } (v_b ^2 - 0)[/tex]
Where [tex]v_b[/tex] is the velocity of the of the ball
Therefore we have from above that
[tex]PE =KE \equiv mg (y_A - y_B) = \frac{1}{2} m (v_b ^2 - 0)[/tex]
Making [tex]v_b[/tex] the subject we have
[tex]v_b = \sqrt{2g (y_A - y_B)}[/tex]
substituting values
[tex]v_b = \sqrt{2g (1.5 - 0.40)}[/tex]
[tex]v_b = 4.6 \ m/s[/tex]
Considering velocity of the ball when it hits the floor in terms of its vertical and horizontal component we have
[tex]v_x = 4.6 m/s \ while \ v_y = 0 m/s[/tex]
The time taken to travel vertically from the point the ball broke loose can be obtained using the equation of motion
[tex]s = v_y t - \frac{1}{2} g t^2[/tex]
Where s is distance traveled vertically which given in the diagram as [tex]s = -0.4[/tex]
The negative sign is because it is moving downward
Substituting values
[tex]-0.4 = 0 -\frac{1}{2} * 9.8 * t^2[/tex]
solving for t we have
[tex]t = 0.3 \ sec[/tex]
Now the distance traveled on the horizontal is mathematically evaluated as
[tex]D = v_b * t[/tex]
[tex]D = 4.6 * 0.3[/tex]
[tex]D = 1.38 m[/tex]
