Question

When an object is thrown from a height of 20 meters with an initial speed of 20 meters per second, its height ℎ above the ground (measured in meters) at any given time (measured in seconds) can be modeled by the quadratic equation: ℎ=−52+20+20 If it is known that the object will reach a maximum height of 40 meters, find the time at which this happens. The object will reach its maximum height of 40 meters at seconds. How long will it take for the object to hit the ground? The object will hit the ground at seconds.

Answer 1

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:

$\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-5)}}{2*1}$

$\dfrac{5 \pm \sqrt{25 +20}}{2}$

$\dfrac{5 \pm \sqrt{45}}{2}$

$\dfrac{5 \pm 6.7082}{2}$

$\dfrac{5 + 6.7082}{2} \ \ OR \ \ \dfrac{5 - 6.7082}{2}$

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.