Projectile motion of ball 2

Classical Mechanics Level 3

A ball tied to the end of a string is in a uniform circular motion at a speed of $v,$ along the circular path with radius $r$ centered at point $O,$ as shown in the above diagram. The height of $O$ from the ground is $2r.$ Suppose that, when the angle between the string and the line drawn from $O$ perpendicularly to the ground is $45°,$ the string is broken loose and the ball flies as indicated by the green dotted line. What is the highest height, $h,$ of the ball during the flight?

$g$ denotes gravitational acceleration.

\frac{v^2}{2g}+\left(\frac{4-\sqrt{2}}{4}\right)r

\frac{v^2}{4g}+\left(\frac{4-\sqrt{2}}{4}\right)r

\frac{v^2}{4g}+\left(\frac{4-\sqrt{2}}{2}\right)r

\frac{v^2}{2g}+\left(\frac{4-\sqrt{2}}{2}\right)r

1 solution

Bhavesh Mendhekar
Mar 28, 2014

max. height reached by the projectile after the breakage of string will be

v^2/4g ( using Hmax = u^2sin^A/2g )

Height of projectile during breakage of string = 2r - r/2^1/2

hence, max. height reached = v^2/4g + (4-2^1/2 / 2)r

where is come frome "2r -r/2^1/2" ? i not understend

Stevanus Setianto - 7 years, 2 months ago

when the string breaks, the object is not at the lowest possible position, it forms an angle of 45 degrees with the vertical, meaning the initial height (where circular motion ends and projectile motion begins) is equal to the total height (2r) minus rcos(45), which results in 2r-r/square root 2.

A Former Brilliant Member - 7 years, 2 months ago

thank you for the explaination

Stevanus Setianto - 7 years, 2 months ago

Projectile motion of ball 2

1 solution

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