I need a math genius please

!!WILL MARK BRAINLIEST!! PLEASE SOMEONE HELP!! IM TRYING TO PREPARE FOR MATH CAMP AND THEY GAVE US THIS PROBLEM AND I CANT FIND ANYTHING LIKE IT!!

I need a math genius pleaseWILL MARK BRAINLIEST PLEASE SOMEONE HELP IM TRYING TO PREPARE FOR MATH CAMP AND THEY GAVE US THIS PROBLEM AND I CANT FIND ANYTHING LI class=

Respuesta :

Answer:

How long is the [shirt] in the air? 3 seconds

How many seconds after launching is the t-shirt at 17 feet? 0.25 seconds

Step-by-step explanation:

Formula to represent the shirt's flight path (given): [tex]h=-16t^2+vt+c[/tex], where [tex]h[/tex] is the height of the shirt, [tex]v[/tex] is the initial velocity of the shirt, [tex]c[/tex] is the shirt's starting height, and [tex]t[/tex] is elapsed time since launch.

The function forms a parabola concave down. Since the shirt is caught at 17 feet, we want to second x-coordinate of a point with a y-coordinate of 17 that the function passes through. This is because the shirt was caught going down, not up.

Therefore, let [tex]h=17[/tex]:

[tex]17=-16t^2+52t+5,\\\\-16t^2+52t-12=0,\\\\ y= \frac{-52\pm\sqrt{52^2-4(-16)(-12)}}{2(-16)},\\\\y=\frac{1}{4},\boxed{y=3}[/tex].

The second x-coordinate is the larger of the two and therefore the shirt was in the air for 3 seconds.

However, the first time the shirt reaches a height of 17 feet is on its way up, which occurs at 1/4 or 0.25 seconds (the first x-coordinate). Therefore, the t-shirt reached a height of 17 feet 0.25 seconds after launching.