Question

Your friend competes in a high jump competition at a track meet. The function h= -16t^2 + 18t models the height h (in feet) of your friend after t seconds. After how many seconds is your friend at a height of 4 feet? After how many seconds does your friend land on the ground?

Answer 1

Answer:

Time when the height is 4 feet is given by

$t= 0.82$ sec or $t=0.30$sec

and time taken to reach the ground is 1.125 sec

Step-by-step explanation:

It is given that the height is given by

$h= -16t^2 +18t$

now we need to find the time t when height h= 4 feet , so we plug h= 4 and solve for t, so we have

$4=-16t^2 +18t$

$16t^2-18t+4=0$  (adding $16t^2$ and subtracting $18t$ to both sides)

now we can use quadratic formula

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

we compare $16t^2-18t+4=0$ with $at^2+bt+c=0$

we have $a=16 , b=-18, c=4$

Now we have

$t=\frac{-(-18)\pm\sqrt{(-18)^2-4(16)(4)} }{2(16)}$

$t=\frac{18\pm\sqrt{68} }{32}$

$t=\frac{18\pm8.246 }{32}$

$t=\frac{18+8.246}{32}$ or $t=\frac{18-8.246}{32}$

so the time when the height is 4 feet is given by

$t= 0.82$ sec or $t=0.30$sec

Now height is 0 on the ground, so we plug h=0 to find t

$0=-16t^2 +18t$

$16t^2 -18t=0$( we bring all the terms to left side)

$2t(8t-9)=0$

$t=0$ or $8t-9=0$ ( we equate each factor to 0)

$t=0$  or $t=\frac{9}{8} = 1.125$

t=0 is the initial time

hence time taken to reach the ground is 1.125 sec

Answer 2

I have attached all my work below. I hope it all makes sense. If not, or if you still have questions feel free to comment below! :)