Question

a ball is thrown directly upward from a height of 7 ft with an initial velocity of 20 ft/sec. the function s(t) = -16t² + 20t + 7 gives the height of the ball, in feet, t seconds after it has been thrown.
determine the time at which the ball reaches its maximum height and find the maximum height.
the ball reaches its maximum height of □ ft sec(s) after the ball is thrown.
(type integers or decimals.)

Explanation:

Step1: Find the time at which the ball reaches its maximum height

The function for the height of the ball is \(s(t)=-16t^{2}+20t + 7\). The formula for the \(t -\)value of the vertex of a quadratic function \(y = ax^{2}+bx + c\) is \(t=-\frac{b}{2a}\).
For \(s(t)=-16t^{2}+20t + 7\), where \(a=-16\) and \(b = 20\).

$$t=-\frac{20}{2\times(-16)}=\frac{20}{32}=\frac{5}{8}=0.625$$

Step2: Find the maximum height

Substitute \(t = 0.625\) into the function \(s(t)\).

$$s(0.625)=-16\times(0.625)^{2}+20\times0.625 + 7$$

$$=-16\times\frac{25}{64}+12.5 + 7$$

$$=-\frac{25}{4}+12.5 + 7$$

$$=-6.25+12.5 + 7$$

$$=13.25$$

Answer:

The ball reaches its maximum height at \(t = 0.625\) seconds and the maximum height is \(13.25\) feet.