Answer:
10.47 s
Step-by-step explanation:
We must solve the quadratic equations for t.
1. On the moon
h = -0.8t² + 10t +2
The standard form of a quadratic equation is
y = ax² + bx + c = 0
By comparing like terms, we see that
a = -0.8; b = 10; c = 2
We can now use the quadratic formula:
[tex]x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{-b\pm\sqrt{D}}{2a}[/tex]
(a) Evaluate the discriminant D
D = b² - 4ac = 10² - 4 × (-0.8) × 2 = 100 + 6.4 = 106.4
(b) Solve for x
[tex]\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-10\pm\sqrt{106.4}}{2\times(-0.8)}\\\\ & = & \dfrac{-10\pm10.32}{-1.6}\\\\x = \dfrac{-10+10.32}{-1.6}&\qquad& x = \dfrac{-10-10.32}{-1.6}\\\\x = -0.197&\qquad& x = 12.70\\\\\end{array}\\\text{On the moon, the ball will stay in flight for 12.70 s}[/tex]
2. On the Earth
(a) Evaluate the discriminant
D = b² - 4ac = 10² - 4 × (-4.9) × 2 = 100 + 39.2 = 139.2
(b) Solve for x
[tex]\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-10\pm\sqrt{139.24}}{2\times(-4.9)}\\\\ & = & \dfrac{-10\pm11.80}{-9.8}\\\\x = \dfrac{-10 +11.80}{-9.8}&\qquad& x = \dfrac{-10 - 11.80}{-9.8}\\\\x = -0.184&\qquad& x = 2.22 \\\\\end{array}\\\text{On Earth, the ball will stay in the air for 2.22 s}[/tex]
3. Difference in time of flight
Time on Moon - time on Earth = 12.70 s - 2.22 s = 10.47 s
The ball will stay in flight 10.47 s longer on the moon than on Earth.
The graphs below show the times of flight on the Moon and on Earth
