Question

To catch a frisbee, a dog leaps into air with an initial velocity of 14 feet per second. Write a model for the height of the dog above the ground. After how many seconds does the dog land on the ground?

Answer 1

Answer:

Model: $y = 14t+\frac{1}{2}(32.16)t^{2}$

The dog land on the ground after 0.8706 seconds.

Step-by-step explanation:

The model that said the height of the dog above the ground is a model of uniform acceleration motion, so it is:

$y = y_0+v_0t+\frac{1}{2}gt^{2}$

Where $y_0$ is the initial height, $v_0$ is the initial velocity and $g$ is the gravity.

So, replacing $y_0$ by zero, $v_0$ by 14 feet per second and $g$ by $32.16 ft/s^2$, we get that the model  for the height of the dog above the ground is:

$y = 14t+\frac{1}{2}(32.16)t^{2}$

Then, the time t when the dog land on the ground is calculated replacing $y$ by zero and solving for t as:

$y = 14t+\frac{1}{2}(32.16)t^{2}\\0 = 14t+\frac{1}{2}(32.16)t^{2}\\0 =t (14 - 16.08t)$

Now, we have two possible solutions:

$t=0$ or $14-16.08t=0$

Taking into account that t=0 is the time when the dog leaps into air, the time when the dog land on the ground is equal to 0.8706 seconds and it is calculated as:

$14-16.08t=0\\14=16.08t\\\frac{14}{16.08}=t\\t=0.8706$