Q.A ball is held in the position shown with string of length 1 m just taut and then projected horizontally with a velocity of 3 m/s. If the string…

Q.A ball is held in the position shown with string of length 1 m just taut and then projected horizontally with a velocity of 3 m/s. If the string…

Q.A ball is held in the position shown with string of length 1 m just taut and then projected horizontally with a velocity of 3 m/s. If the string…

Q.A ball is held in the position shown with string of length 1 m just taut and then projected horizontally with a velocity of 3 m/s. If the string…
The ball is held in position as shown in the attached image with string of length 1 m just taut and then projected horizontally with a velocity of 3 m/s.Let the angle made by the string with the vertical with it is projected be `theta` . The height of the ball (with reference to a point 1 m below where the string is attached) is `1*cos theta` . The time taken by the ball to drop by `1 – cos theta` is the same as that taken by it to travel a distance `1*sin theta` .This gives `(1 – cos theta) = (1/2)*10*((sin theta)/3)^2`=> `(1 – cos theta) = 5*(sin^2 theta)/9`=> `(1 – cos theta) = 5*(1 – cos^2 theta)/9` => `9/5 = 1 + cos theta` => `cos theta = 9/5 – 1 = 4/5` => `theta = cos^-1(4/5)` => `theta` = 37 degrees