Respuesta :
Answer:
t = 5.25 seconds
Step-by-step explanation:
Given that,
A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation:
[tex]h(t)=-16t^2+48t+190[/tex]
It is assumed to find the time when the projectile will hit the ground. When the projectile hit the ground, its height is equal to 0 such that,
[tex]-16t^2+48t+190=0[/tex]
It forms a quadratic equation such that,
[tex]t=\dfrac{-b+\sqrt{b^{2}-4ac } }{2a},\dfrac{-b-\sqrt{b^{2}-4ac } }{2a}\\\\t=\dfrac{-48+\sqrt{(48)^{2}-4\times (-16)(190) } }{2\times (-16)},\dfrac{-48-\sqrt{(48)^{2}-4\times (-16)(190) } }{2\times (-16)}\\\\t=-2.25, 5.25[/tex]
So, the projectile will hit the ground at t = 5.25 seconds.
