# 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 becomes taut again when it is vertical, angle...

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 becomes taut again when it is vertical, angle `theta` is given by:

A) 53°

B) 30°

C) 45°

D) 37°

(cos 37° = 4/5 , cos 53° = 3/5)

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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