Question

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) = –16t2 + 48t + 190.

Answer 1

Answer: C

Step-by-step explanation: took the test

Answer 2

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,

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

So, the projectile will hit the ground at t = 5.25 seconds.