Given:
A ball is kicked 4 feet above the ground with an initial vertical velocity of 55 feet per second.
The function [tex]h(t)=-16t^2+55t+4[/tex] represents the height h(in feet) of the ball after t seconds.
We need to determine the time of the ball at which it is 30 feet above the ground.
Time:
To determine the time that it takes for the ball to reach a height of 30 feet above the ground, let us substitute h(t) = 30, we get;
[tex]30=-16t^2+55t+4[/tex]
[tex]26=-16t^2+55t[/tex]
Adding both sides of the equation by 16t², we get;
[tex]16t^2+26=55t[/tex]
Subtracting both sides of the equation by 55t, we have;
[tex]16t^2-55t+26=0[/tex]
Let us solve the quadratic equation using the quadratic formula, we get;
[tex]t=\frac{-(-55) \pm \sqrt{(-55)^2-4(16)(26)}}{2(16)}[/tex]
[tex]t=\frac{55 \pm \sqrt{3025-1664}}{32}[/tex]
[tex]t=\frac{55 \pm \sqrt{1361}}{32}[/tex]
[tex]t=\frac{55 \pm 36.89}{32}[/tex]
[tex]t=\frac{55 + 36.89}{32} \ or \ t=\frac{55- 36.89}{32}[/tex]
[tex]t=\frac{91.89}{32} \ or \ t=\frac{18.11}{32}[/tex]
[tex]t=2.9 \ or \ t=0.6[/tex]
The value of t is t = 0.6 because this denotes the time taken by the ball to reach a height of 30 feet from the ground.
Therefore, the time taken by the ball to reach a height of 30 feet above the ground is 0.6 seconds.
