Question

An astronaut on the Moon throws a baseball upward with an initial velocity of 10 meters per second, letting go of the baseball 2 meter above the ground. The equation of the baseball pathway can be modeled by h=-0.8t^2+10t+2. The same experiment is done on Earth, in which the pathway is modeled by the equation h=-4.9t^2+10t+2. How much longer would the ball stay in the air on the Moon compared to on Earth

Answer 1

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:

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

(a) Evaluate the discriminant D

D = b² - 4ac = 10² - 4 × (-0.8) × 2 = 100 + 6.4  = 106.4

(b) Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-10\pm\sqrt{106.4}}{2\times(-0.8)}\\\\ & = & \dfrac{-10\pm10.32}{-1.6}\\\\x = \dfrac{-10+10.32}{-1.6}&\qquad& x = \dfrac{-10-10.32}{-1.6}\\\\x = -0.197&\qquad& x = 12.70\\\\\end{array}\\\text{On the moon, the ball will stay in flight for 12.70 s}$

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

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-10\pm\sqrt{139.24}}{2\times(-4.9)}\\\\ & = & \dfrac{-10\pm11.80}{-9.8}\\\\x = \dfrac{-10 +11.80}{-9.8}&\qquad& x = \dfrac{-10 - 11.80}{-9.8}\\\\x = -0.184&\qquad& x = 2.22 \\\\\end{array}\\\text{On Earth, the ball will stay in the air for 2.22 s}$

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