Question

the height s of a ball t seconds when thrown straight up with an initial speed of 75 feet per second from an initial height of 5 feet (not reached by the ball) is given by (s(t)= - 16t^{2}+75t + 5) (a) when will the height of the ball be 40 feet? (b) when will the height of the ball be 70 feet? (c) will the ball ever reach a height of 130 feet? (a) the height of the ball will be 40 feet after approximately seconds. (use a comma to separate answers as needed. round to the nearest tenth as needed.) (b) the height of the ball will be 70 feet after approximately seconds. (use a comma to separate answers as needed. round to the nearest tenth as needed.)

Explanation:

Step1: Identify the height - time formula

Let the height - time formula be $s(t)=- 16t^{2}+v_{0}t + s_{0}$, where $v_{0}$ is the initial velocity and $s_{0}$ is the initial height. Since the initial height $s_{0} = 5$ feet and initial velocity $v_{0}=75$ feet per second, the formula is $s(t)=-16t^{2}+75t + 5$.

Step2: Solve for part (a)

Set $s(t)=40$, so we have the quadratic equation $-16t^{2}+75t + 5=40$, which simplifies to $-16t^{2}+75t - 35 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a=-16$, $b = 75$, and $c=-35$. Then $t=\frac{-75\pm\sqrt{75^{2}-4\times(-16)\times(-35)}}{2\times(-16)}=\frac{-75\pm\sqrt{5625 - 2240}}{-32}=\frac{-75\pm\sqrt{3385}}{-32}=\frac{-75\pm58.18}{-32}$. We get two solutions: $t_{1}=\frac{-75 + 58.18}{-32}=\frac{-16.82}{-32}\approx0.5$ and $t_{2}=\frac{-75 - 58.18}{-32}=\frac{-133.18}{-32}\approx4.2$.

Step3: Solve for part (b)

Set $s(t)=70$, so $-16t^{2}+75t + 5=70$, which simplifies to $-16t^{2}+75t - 65 = 0$. Using the quadratic formula with $a=-16$, $b = 75$, and $c=-65$. $t=\frac{-75\pm\sqrt{75^{2}-4\times(-16)\times(-65)}}{2\times(-16)}=\frac{-75\pm\sqrt{5625-4160}}{-32}=\frac{-75\pm\sqrt{1465}}{-32}=\frac{-75\pm38.27}{-32}$. We get $t_{1}=\frac{-75 + 38.27}{-32}=\frac{-36.73}{-32}\approx1.1$ and $t_{2}=\frac{-75 - 38.27}{-32}=\frac{-113.27}{-32}\approx3.5$.

Step4: Solve for part (c)

Set $s(t)=130$, so $-16t^{2}+75t + 5=130$, which simplifies to $-16t^{2}+75t - 125 = 0$. Calculate the discriminant $\Delta=b^{2}-4ac$, where $a=-16$, $b = 75$, and $c=-125$. $\Delta=75^{2}-4\times(-16)\times(-125)=5625 - 8000=-2375<0$. Since the discriminant is negative, the ball will never reach a height of 130 feet.

Answer:

(a) The height of the ball will be 40 feet after approximately 0.5 seconds and 4.2 seconds.
(b) The height of the ball will be 70 feet after approximately 1.1 seconds and 3.5 seconds.
(c) No, the ball will never reach a height of 130 feet.