if the string is released at the instant the ball is at the top of the loop, 139.23 cm far to the right does the ball hit the ground.
A minimum velocity is present at a position where acceleration is both negative to the left and positive to the right.
At the minimum velocity at the top v, the weight mg provides the centripetal acceleration -
R = 50 cm = 0.5 m
Thus, mv²/R = mg
or, v² = Rg
or, v = v Rg
or, v = v × 0.50 x 9.8
or, v =v × 4.9
or, v = 2.21 m/s
The ball goes off horizontally at this speed and falls through a height
y = h + R
or, y = 150 + 50
or, y = 200 cm
or, y = 2 m
Using y = 0 + 0.5g.t2
or, t = v (2.y/g)
or, t = v [2 x 2/9.8]
or, t = 0.63 s
The horizontal distance travelled, x = v.t
x = 2.21 x 0.63
x = 1.3923 m
x = 139.23 cm.
If the string is released at the instant the ball is at the top of the loop, 139.23 cm far to the right does the ball hit the ground.
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