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. approximately when will the projectile hit the ground? 1.5 seconds 3.2 seconds 5.3 seconds 6.2 seconds

Answer 1

Answer:

5.3 seconds

Step-by-step explanation:

We have the motion equation as follows

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

When the projectile touches the ground we have the height equal to 0

-16t^{2}+ 48t +190=0

we need to to solve this quadratic equation with:

$\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

$\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

In this case A=-16, B=48, C=190

$\frac{-48+\sqrt{48^{2}-4*(-16)190 } }{2(-16)}$

$\frac{-48-\sqrt{48^{2}-4*(-16)190 } }{2(-16)}$

The results are -2.25 s and 5.3 s, since time is a positive variable the final answer is 5.3 seconds

Answer 2

I just graphed this equation, and it shows that the projectile will hit the ground at 5.3 seconds. I could be wrong but I'm going with the calculator is right I hope. I hope this helps you!