A football is kicked from the ground and follows the graph of the function y = –t2 + 3t, where y represents height and t represents seconds. At the same time, a drone is launched from a height of 4 feet and follows the straight path of y = –t + 4. At what time are the football and the drone at the same height?

Options:
A. 2 seconds
B. 3 seconds
C. 1 second
D. 5 seconds

Respuesta :

Answer:

The height of the objects are the same after 2 seconds.

Step-by-step explanation:

In order to calculate at which time both objects have the same height we need to find the value of t that makes both equations equal. Therefore:

[tex]-t^2 + 3t = -t + 4\\t^2 - 3t - t + 4 = 0\\t^2 - 4t + 4 = 0\\t_{1,2} = \frac{-(-4) \pm \sqrt{(-4)^2 - 4*1*4}}{2*1} = \frac{4 \pm \sqrt{16 - 16}}{2}\\t_{1,2} = \frac{4}{2} = 2[/tex]

The height of the objects are the same after 2 seconds.