An astronaut on the moon throws a ball upward. The astronaut releases the ball at a height of 6 feet (ho). with an initial velocity (vo) of 20 feet/second. The height (h) of the ball in feet is given by the equation:
h(t) = -2.7t^2 + vot + ho
Note: t represents the number of seconds after the ball was thrown.
a) What is the amount of time it takes the ball to achieve a height of 25 feet above the moon's surface?
b) At what height will the ball be after 5 seconds?
c) How long does it take for the ball to return to the surface of the moon?
don't forget your Algebra I now that you're taking calculus ...
(a) just solve for t in -2.7t^2 + 20t + 6 = 25
(b) just find h(5)
(c) solve for t in -2.7t^2 + 20t + 6 = 0
a) To find the amount of time it takes the ball to achieve a height of 25 feet above the moon's surface, we need to solve the equation h(t) = 25 for t.
h(t) = -2.7t^2 + vt + h0 (where h(t) represents the height at time t, v represents the initial velocity, and h0 represents the initial height)
Substituting the given values:
25 = -2.7t^2 + 20t + 6
Now, let's rearrange the equation to solve for t:
2.7t^2 - 20t + 19 = 0
This is a quadratic equation, and we can solve it by factoring, completing the square, or using the quadratic formula.
By solving this equation, we find two possible values for t: t ≈ 1.975 seconds and t ≈ 6.792 seconds.
Therefore, the amount of time it takes the ball to achieve a height of 25 feet above the moon's surface is approximately 1.975 seconds and 6.792 seconds.
b) To find the height of the ball after 5 seconds, we can substitute t = 5 into the equation h(t) = -2.7t^2 + vt + h0.
h(5) = -2.7(5)^2 + 20(5) + 6
h(5) = -2.7(25) + 100 + 6
h(5) = -67.5 + 100 + 6
h(5) = 38.5
Therefore, after 5 seconds, the ball will be at a height of approximately 38.5 feet above the moon's surface.
c) To find the time it takes for the ball to return to the surface of the moon, we need to solve the equation h(t) = 0 for t.
h(t) = -2.7t^2 + vt + h0
Setting h(t) = 0, we have:
0 = -2.7t^2 + 20t + 6
Again, we need to solve this quadratic equation. By solving it, we find two possible values for t: t ≈ 0.484 seconds and t ≈ 7.601 seconds.
Therefore, it takes approximately 0.484 seconds and 7.601 seconds for the ball to return to the surface of the moon.
To answer these questions, we need to substitute the given values into the equation h(t) = -2.7t^2 + vot + ho and solve for the different unknowns.
a) What is the amount of time it takes the ball to achieve a height of 25 feet above the moon's surface?
To find the time it takes for the ball to achieve a height of 25 feet, we set h(t) equal to 25 and solve for t.
h(t) = -2.7t^2 + vot + ho
25 = -2.7t^2 + 20t + 6
Now, we have a quadratic equation in the form ax^2 + bx + c = 0, where a = -2.7, b = 20, and c = 6 - 25 = -19.
To solve this quadratic equation, we can either use the quadratic formula or factorize. Let's use the quadratic formula.
The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values, we get:
t = (-20 ± √(20^2 - 4(-2.7)(-19))) / (2(-2.7))
Simplifying further:
t = (-20 ± √(400 - 205.2)) / (-5.4)
t = (-20 ± √(194.8)) / (-5.4)
Since time cannot be negative, we'll consider only the positive square root:
t ≈ (-20 + 13.97) / (-5.4) ≈ -6.03 / (-5.4) ≈ 1.12 seconds.
Therefore, it takes approximately 1.12 seconds for the ball to achieve a height of 25 feet above the moon's surface.
b) At what height will the ball be after 5 seconds?
To find the height of the ball after 5 seconds, we substitute t = 5 into the equation.
h(t) = -2.7t^2 + vot + ho
h(5) = -2.7(5)^2 + 20(5) + 6
h(5) = -2.7(25) + 100 + 6
h(5) = -67.5 + 100 + 6
h(5) = 38.5
Therefore, the ball will be at a height of 38.5 feet after 5 seconds.
c) How long does it take for the ball to return to the surface of the moon?
To find the time it takes for the ball to return to the surface, we need to find when the height h(t) becomes 0.
h(t) = -2.7t^2 + vot + ho
0 = -2.7t^2 + 20t + 6
Setting h(t) to 0, we have another quadratic equation to solve. We can use the quadratic formula or factorize.
Let's use the quadratic formula again:
t = (-20 ± √(20^2 - 4(-2.7)(6))) / (2(-2.7))
Simplifying further:
t = (-20 ± √(400 + 64.8)) / (-5.4)
t = (-20 ± √(464.8)) / (-5.4)
Again, we'll consider only the positive square root:
t ≈ (-20 + √(464.8)) / (-5.4) ≈ (-20 + 21.55) / (-5.4) ≈ 1.41 / (-5.4) ≈ -0.26 seconds.
Since time cannot be negative, we discard this solution.
Therefore, the ball does not return to the surface of the moon in the given time frame.