3 years ago

I need help fast. 8th grade multiple question. (I added photos)

Mathematics

1 answer:

n200080 [17]3 years ago

5 0

Answer:

C is the answer

Step-by-step explanation:

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3 years ago

A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectil

Furkat [3]

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:

$h(t)=-16t^2+48t+190$

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,

$-16t^2+48t+190=0$

It forms a quadratic equation such that,

$t=\dfrac{-b+\sqrt{b^{2}-4ac } }{2a},\dfrac{-b-\sqrt{b^{2}-4ac } }{2a}\\\\t=\dfrac{-48+\sqrt{(48)^{2}-4\times (-16)(190) } }{2\times (-16)},\dfrac{-48-\sqrt{(48)^{2}-4\times (-16)(190) } }{2\times (-16)}\\\\t=-2.25, 5.25$