Answer:
70 seconds
Step-by-step explanation:
Given that;
The initial velocity [tex]v_o[/tex] of the bullet fired = 1120 ft/s
Initial height h = 8 feet
The expression to determine how many seconds it takes the bullet to hit the ground is:
[tex]h = -16t^2 +v_ot + 8[/tex]
Thus;
Replacing the value of [tex]v_o[/tex] = 1120 and h = 0 (i.e. when h =0) in the above expression; we have:
[tex]0= -16t^2 +(1120)t + 8[/tex]
[tex]= -16t^2 +(1120)t + (8-0)[/tex]
= -16t² + 1120t + 8
mulitiply through by (-)
= 16t² -1120t - 8
Divide through by 8
= 2t² - 140t - 1
The above expression forms a quadratic equation.
where;
a = 2
b = -140
c = - 1
So, by using the quadratic formula [tex]\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex], we have:
[tex]= \dfrac{-(-140) \pm \sqrt{(-140)^2-4(2)(-1)}}{2(2)}[/tex]
[tex]= \dfrac{140 \pm \sqrt{19600-(-8)}}{4}[/tex]
[tex]= \dfrac{140 \pm \sqrt{19608}}{4}[/tex]
[tex]= \dfrac{140 \pm 140}{4}[/tex]
[tex]=\dfrac{140+ 140}{4} \ \ \ OR \ \ \ \dfrac{140-140}{4}[/tex]
[tex]=\dfrac{280}{4} \ \ \ OR \ \ \ 0[/tex]
= 70
Thus, the time (in seconds) it took the bullet to it the ground = 70 seconds