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?

Answer 1

The projectile hit the ground approximately at t = 5.3 seconds.

It is given that projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall.

It is required to find when will the projectile hit the ground if the projectile equation is $\rm h(t)= -16t^2+48t+190$

What is a parabola?

It is defined as the graph of a quadratic function that has something bowl-shaped.

We have the projectile equation:

$\rm h(t)= -16t^2+48t+190$ .....(1)

When it hit the ground the value of h(t) = 0, equate the equation(1) with 0

We get:

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

or

$\rm 16t^2-48t-190= 0$

It is a quadratic equation,

Compare with the equation $ax^2+bx+c=0$

a = 16, b = -48, and c = -190

By using the formula:

$\rm x= \frac{-b\pm\sqrt{b^2-4ac} }{2a}$

$\rm t= \frac{-(-48)\pm\sqrt{(-48)^2-4(16)(-190)} }{2(16)}\\\\\rm t= \frac{48\pm{(120.266)} }{32}$

t = -2.25  or t = 5.25

Time can not be a negative quantity, so leave the negative value of 't'.

So, t = 5.25 ≈ 5.3 seconds

Thus, the projectile hit the ground at t = 5.3 seconds.

Know more about the parabola here:

brainly.com/question/8708520