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?
a. 1.5 secondsB. 3.2 secondsC. 5.3 secondsD. 6.2 seconds

Answer 1

For this case we have the following quadratic equation:
 $h (t) = -16t ^ 2 + 48t + 190$
 By the time the projectile hits the ground we have the final height equal to zero:
 $-16t ^ 2 + 48t + 190 = 0$
 Therefore, we have a second-order polynomial that we must solve:
 Using resolver we have:
 $x = \frac{-48+/-\sqrt{48^2-4(-16)(190)} }{2(-16)}$
 Rewriting:
 $x = \frac{-48+/-\sqrt{2304+12160} }{-32}$
 $x = \frac{-48+/-\sqrt{14464} }{-32}$
 Doing the calculations we have two roots:
 $t1 = 5.3 t2 = -2.3$
 We discard the negative root because we are looking for time which is greater than zero.
 Answer:
 the projectile will hit the ground after:
 C. 5.3 seconds

Answer 2

Answer:

C- 5.3 Seconds