Question

gravitation the height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation h(t)=-16t² +60t. at what times will the object be at a height of 56 feet?

Explanation:

Step1: Set up the equation

We want to find \(t\) when \(h(t)=56\). So we set up the equation \(-16t^{2}+60t = 56\).

Step2: Rearrange the equation

Rearrange it to the standard quadratic form \(ax^{2}+bx + c=0\). We get \(16t^{2}-60t + 56=0\). Divide through by \(4\) to simplify: \(4t^{2}-15t + 14 = 0\).

Step3: Use the quadratic formula

The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For \(a = 4\), \(b=-15\), \(c = 14\), we first calculate the discriminant \(\Delta=b^{2}-4ac=(-15)^{2}-4\times4\times14=225 - 224=1\).

Step4: Calculate \(t\) values

Then \(t=\frac{15\pm\sqrt{1}}{8}\). So \(t_1=\frac{15 + 1}{8}=\frac{16}{8}=2\) and \(t_2=\frac{15-1}{8}=\frac{14}{8}=\frac{7}{4}=1.75\).

Answer:

The object will be at a height of 56 feet at \(t = 1.75\) seconds and \(t = 2\) seconds.