Question

An egg is dropped from the window of a building. The height, in feet, of the egg t seconds after it is thrown is represented by d=-16t^2-7t+61. How many seconds after the egg is thrown will it be 10 feet from the ground?

Answer 1

Answer:

1.58

Step-by-step explanation:

Okay. We're solving for t because we need to know how many seconds. From the problem, we know the height (10), but we don't know when that height happens.

$10 = -16tx^{2} -7t+61$

We can factor by multiplying -16 by 61 first to get -976 and see what multiplies to get -976 AND adds to get -7 at the same time. But we can also use the quadratic formula because, let's be honest, factoring can be a lot of extra work.

$10 = -16t^{2} -7t+61\\\\0 = -16t^{2} -7t+51\\\\t=\frac{-b\pm \sqrt{b^{2}-4ac } }{2a} \\=\frac{-(-7)\pm \sqrt{(-7)^{2}-4(-16)(51) } }{2(-16)} \\=\frac{7\pm \sqrt{49+64(51) } }{-32} \\=\frac{7\pm \sqrt{3313} }{-32}$

Plug both the plus and the minus versions into your calculator and...

$t\approx -2.02, 1.58$

There's no such thing as negative time, so -2.02 is automatically disqualified, leaving 1.58 as the answer.

Hope this helps, let me know if I missed anything!