Answer:
5.85 seconds
Step-by-step explanation:
The equation;
h = -5t² + vt + h
where;
v = initial upward velocity in meter per seconds
h = height above ground level when the object is thrown
t = time of flight in seconds
h = -5t² + 20t + 20
h = 0
⇒ -5t² + 20t + 20= 0
-t² + 5t + 5 = 0
t² - 5t - 5 = 0
a = 1
b = -5
c = -5
Using the quadratic formula:
[tex]\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]\dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-5)}}{2*1}[/tex]
[tex]\dfrac{5 \pm \sqrt{25 +20}}{2}[/tex]
[tex]\dfrac{5 \pm \sqrt{45}}{2}[/tex]
[tex]\dfrac{5 \pm 6.7082}{2}[/tex]
[tex]\dfrac{5 + 6.7082}{2} \ \ OR \ \ \dfrac{5 - 6.7082}{2}[/tex]
t = 5.8541 OR t = -0.8541
Since t can only be positive; t ≅ 5.85 sec
Then;
After 5.85 seconds, the object will hit the ground.
