Question

If an object is propelled upward from a height of 72 feet at an initial velocity of 90 feet per second, then its height h after t seconds is given by the equation h = − 16 t 2 + 90 t + 72 . After how many seconds does the object hit the ground?

Answer 1

Answer:

6.34 seconds.

Step-by-step explanation:

The object will hit the ground when h = 0.

-16t^2 + 90t + 72 = 0

8t^2 - 45t - 36 = 0

We can then use the quadratic formula to solve.

[please ignore the A-hat; that is a bug]

$\frac{45±\sqrt{45^2 - 4 * 8 * -36} }{2 * 8}$

= $\frac{45±\sqrt{2025 + 1152} }{16}$

= $\frac{45±\sqrt{3177} }{16}$

= $\frac{45±56.36488268}{16}$

(45 - 56.36488268) / 16 = -0.7103051678

(45 + 56.36488268) / 16 = 6.335305168

Since the time cannot be negative, the object will hit the ground after about 6.34 seconds.

Hope this helps!