A ball is thrown vertically in the air with a velocity of 95ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in seconds, the ball is at a height of 120ft. Round your answer(s) to the nearest tenth of a second.

Answer:[tex]t_1 = 1.8\ s\\\\t_2 = 4.1\ s[/tex]Step-by-step explanation:If the equation [tex]h = -16t^2 + v_0t[/tex] represents the position of the ball as a function of time then, to find in which second the ball reaches 120 feet must substitute [tex]h = 120[/tex] in the equation of the height and solve for t.[tex]120 = -16t ^ 2 + v_0t[/tex]If the initial velocity is 95 feet/s then [tex]v_0 = 95[/tex]Then:[tex]120 = -16t ^ 2 + 95t\\\\-16t ^ 2 + 95t -120 = 0[/tex]Use the quadratic formula[tex]t_1 = \frac{-b+\sqrt{b^2 -4ac}}{2a}\\\\t_2 = \frac{-b-\sqrt{b^2 -4ac}}{2a}[/tex]Where, for this problem:[tex]a = -16\\b = 95\\c = -120[/tex]So[tex]t_1 =\frac{-95+\sqrt{(95)^2 -4(-16)(-120)}}{2(-16)} = 1.8\ s\\\\t_2=\frac{-95+\sqrt{(95)^2 -4(-16)(-120)}}{2(-16)} = 4.1\ s[/tex]This result means that the ball reaches 120 feet for the first time at 1.8 seconds, then begins to descend and on its descent again reaches 120 feet at t = 4.1 seconds.