Suspended by a 1.5 m long string, a heavy particle is given an initial horizontal velocity of 57 m/s . Determine the maximum height reached by the particle above the point of suspension. Take g = 10 m/s ^2 .

Step-by-step solution

Given: Initial velocity, u = 57 m/s Length of the string, l = 1.5 m Acceleration due to gravity, g = 10 m/s ^2 The minimum velocity required to complete the vertical circle is 5gl = 5 10 1.5 = 75 m/s . The minimum velocity required to reach the horizontal position is 2gl = 2 10 1.5 = 30 m/s . Since 30 Let the string become slack at an angle with the upward vertical. At this point, the tension in the string is zero ( T = 0 ). The component of gravity provides the necessary centripetal force: mg = mv^2 l v^2 = gl Usi