**Problem Description**A ball is thrown vertically upward with an initial speed of 88 ft/sec from the base \(A\) of a cliff with \(H = 60\) ft. Determine the distance \(h\) by which the ball clears the top of the cliff and the time \(t\) after release for the ball to land at \(B\). Also, calculate the magnitude of the impact velocity \(v_B\). Neglect air resistance and the small horizontal motion of the ball.**Diagram Explanation**The diagram illustrates the problem setup:- Point \(A\) is at the base of the cliff.- A ball is thrown upward from point \(A\) with an initial velocity \(v_0 = 88\) ft/sec.- The cliff has a height \(H = 60\) ft.- Point \(B\) is at the top of the cliff.- The distance \(h\) represents the height by which the ball clears the top of the cliff when it reaches its maximum height. **Equations to Use**To solve this problem, we can use the following kinematic equations:1. \(v = u + at\)2. \(s = ut + \frac{1}{2}at^2\)3. \(v^2 = u^2 + 2as\)Where:- \(u\) is the initial velocity (88 ft/sec)- \(v\) is the final velocity- \(a\) is the acceleration due to gravity (\(-32.2\) ft/sec\(^2\))- \(t\) is the time- \(s\) is the distance**Answers**Fill in the answers after solving the equations:\( h = \) ______ ft\( t = \) ______ sec\( v_B = \) ______ ft/sec

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