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 meters 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 equation h= -4.9t^2+10t+2. How much longer would the ball stay in the air on the moon compared to on the Earth

Answer 1

Answer:

The baseball will stay 10.47s longer on the moon than on the earth.

Step-by-step explanation:

The amount of time the baseball on the moon will stay  in air is until

$h(t)= -0.8t^2+10t+2 =0$. (i.e when the ball reaches the ground)

Similarly, the amount of time the baseball on earth will stay in air is until

$h(t)-4.9t^2+10t+2=0$

The solution to these equations can be found using the quadratic formula.

For the baseball on the moon

$-0.8t^2+10t+2 =0$

$t = \dfrac{-10\pm \sqrt{10^2-4(-0.8*2)} }{2*-0.8}$

whose positive solution is

$\boxed{t= 12.697s}$

And for the baseball on earth

$-4.9t^2+10t+2 =0$

$t = \dfrac{-10\pm \sqrt{10^2-4(-4.9*2)} }{2*-4.9}$

whose positive solution is

$\boxed{t = 2.224s}$

Thus, the baseball will stay $12.697s-2.2243s=10.472s$ longer on the moon than on the earth.