Option B
At 2 second and 1.75 second, the object be at a height of 56 feet
Solution:
Given that,
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:
[tex]h(t) = -16t^2 + 60t[/tex]
At what times will the object be at a height of 56 feet
Substitute h = 56
[tex]56 = -16t^2 + 60t\\\\16t^2 - 60t + 56 = 0\\\\Divide\ the\ equation\ by\ 4\\\\4t^2 - 15t + 14=0[/tex]
Solve the above equation by quadratic formula
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\mathrm{For\:}\quad a=4,\:b=-15,\:c=14\\\\x =\frac{-\left(-15\right)\pm \sqrt{\left(-15\right)^2-4\cdot \:4\cdot \:14}}{2\cdot \:4}\\\\Simplify\\\\x = \frac{15 \pm \sqrt{1}}{8}\\\\x = \frac{15 \pm 1}{8}\\\\We\ have\ two\ solutions\\\\x = \frac{15+1}{8} \text{ and } x = \frac{15-1}{8}\\\\x = 2 \text{ and } 1.75[/tex]
Thus, at 2 second and 1.75 second, the object be at a height of 56 feet
