A stone is thrown vertically upwards and its height at any time t seconds is \(h = 45t - 9t^{2}\). Find the maximum height reached.

Assessment: WAEC SSCE - Further Mathematics - 2010 (Objective) Subject: Further Mathematics

Question 1 Report

A stone is thrown vertically upwards and its height at any time t seconds is \(h = 45t - 9t^{2}\). Find the maximum height reached.

Answer Details
The height of the stone at any time t seconds is given by \(h = 45t - 9t^{2}\). To find the maximum height reached, we need to find the vertex of the parabolic function \(h\), since the vertex represents the highest point of the parabola. The vertex of the parabola is given by the formula \(\frac{-b}{2a}\), where \(a\) and \(b\) are the coefficients of the quadratic function. In this case, \(a = -9\) and \(b = 45\). Substituting these values into the formula for the vertex, we get: \[\frac{-b}{2a} = \frac{-45}{2(-9)} = \frac{45}{18} = 2.5.\] Therefore, the stone reaches its maximum height after 2.5 seconds. To find the maximum height, we substitute \(t = 2.5\) into the equation for \(h\): \[h = 45(2.5) - 9(2.5)^2 = 56.25\text{ m}.\] Therefore, the maximum height reached by the stone is 56.25 meters. The answer is (D) 56.25 m.

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