Question

coach roberts has a device that launches baseballs into the air so the outfielders can practice catching pop - ups. if it launches a ball with a velocity of 120 feet per second, after how many seconds will the ball be at a height of 130 feet above the ground? use the equation for projectile motion ( h=-16t^{2}+v_{0}t ) where ( h ) is the height of the ball ( t ) seconds after it is launched at an initial velocity of ( v_{0} ). enter multiple answers separated by a comma and round all of your answers to one decimal place.

Explanation:

Step1: Substitute values into the equation

Given \(h = 130\), \(v_0=120\), substitute into \(h=-16t^{2}+v_0t\).
We get \(130=-16t^{2}+120t\).
Rearrange to standard quadratic form \(16t^{2}-120t + 130=0\). Divide by 2: \(8t^{2}-60t + 65=0\).
Here \(a = 8\), \(b=-60\), \(c = 65\).

Step2: Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

First, calculate the discriminant \(\Delta=b^{2}-4ac=(-60)^{2}-4\times8\times65=3600 - 2080=1520\).
Then \(t=\frac{60\pm\sqrt{1520}}{16}\).
Since \(\sqrt{1520}\approx38.99\), \(t=\frac{60\pm38.99}{16}\).
For the plus - case: \(t=\frac{60 + 38.99}{16}=\frac{98.99}{16}\approx6.2\).
For the minus - case: \(t=\frac{60-38.99}{16}=\frac{21.01}{16}\approx1.3\).

Answer:

\(1.3,6.2\)