Question

An object is thrown upward at a speed of 64 feet per second by a machine from a height of 16 feet off the ground. The height h of the object after t seconds can be found using the equation h=−16t^2+64t+16When will the height be 73 feet?Round to two places.

Answer 1

Answer:

0.75 second

Explanation:

The equation is

$h = 16t^{2}+64 t + 16$

Let t be the time when it is at 73 feet. Substitute, h = 73 in the above equation and then find the value of t.

$73 = 16t^{2}+64 t + 16$

$16t^{2}+64 t-57=0$

$t=\frac{-64\pm \sqrt{64^{2}+4\times 16\times 57}}{2\times 16}$

$t=\frac{-64\pm 88}{32}$

Take positive sign

t = 0.75 second