Question

1. An astronaut standing on the moon's surface throws a rock upward with an initial velocity of 50 feet per second. The height of the rock can be modeled by m = -2.7t² + 50t+ 6, where m is the height of the rock (in feet) and t is the time (in seconds). a. How high will the rock go? b. How long will it take the rock to hit the ground? C. If the astronaut throws the same rock upward with the same initial velocity on Earth, the height of the rock is modeled by e = = -16t² + 50t + 6. Would the rock hit the ground in less time on the moon or on Earth? Explain your answer.​

Answer 1

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Explanation:

a. To find the maximum height of the rock, we need to find the vertex of the function. The vertex occurs at t = -b/2a, where a = -2.7 and b = 50. So, t = -50 / (2*(-2.7)) = 9.26 seconds. Now we can substitute this value into the equation to find the maximum height:

m = -2.7(9.26)² + 50(9.26) + 6 = 122.2 feet

So the rock will go up to a height of 122.2 feet.

b. To find when the rock will hit the ground, we need to find the time when the height of the rock is 0. So, we can set m = 0 and solve for t:

-2.7t² + 50t + 6 = 0

Using the quadratic formula, we get:

t = (-50 ± sqrt(50² - 4*(-2.7)*6)) / (2*(-2.7))

t = 18.52 seconds or t = 0.21 seconds

Since the negative root doesn't make sense in this context, we can assume that the rock hits the ground after 18.52 seconds.

c. The time it takes for the rock to hit the ground on Earth is given by setting e = 0:

-16t² + 50t + 6 = 0

Using the quadratic formula, we get:

t = (-50 ± sqrt(50² - 4*(-16)*6)) / (2*(-16))

t = 3.71 seconds or t = 0.10 seconds

So, the rock would hit the ground much faster on Earth than on the moon. This is because the gravitational force on the moon is much weaker than on Earth, so it takes longer for objects to fall to the ground.