Suppose a stone is thrown vertically upward from the edge of a...

Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 64 ft/s from a height of 32 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t²+64t+32.
c. What is the height of the stone at the highest point?

c. What is the height of the stone at the highest point?

Accepted Answer

Welcome back, everyone. A basketball is shot upwards from the ground with an initial velocity of 40 ft per second. The height and feet of the basketball T seconds after this shot is given by the equation H of T equals -16 T2 + 40T. What is the maximum height reached by the basketball? We're going for an answer choices AS 20 ft, B 25 ft, C 35 ft, and D 40 ft. So let's analyze the height function we're given H of T, and we want to find its maximum. So we're going to identify the first derivative of the function. So let's differentiate minus 16 T2 plus 40 T. And for this purpose we're going to apply the power rule. So the derivative of T2d is TT, meaning we get negative 302 T, and the derivative of T is 1. So we're just adding 40 multiplied by 1, which is 40, and we want to identify the critical points. So we're going to set this derivative equal to 0. Which essentially means that 32 T + 40. Equals 0 we can factor out. 8, and this gives us, or T? Minus 5 equals 0. So we can say that 4 T minus 5 is equal to 0. And now 4 T equals 5, so that time T equals 5 divided by 4. What we want to do is basically verify that this is the point of maximum. And we can change, we can check. The sign of the derivative. So, first of all, let's identify the sign of the derivative when time is less than the critical point, so less than 5/4s. And we can take any value within the domain, time. is non-negative, so we can take H at 0.0. Let's substitute 04 T. And this gives us a -32 multiplied by 0 + 40, so it's 40. And this gives us a positive value, meaning the function is increasing, right? Our height is increasing. Now let's valid each prime for time values greater than 5/4, let's oppose 2. So H at. 2 is equal to -32 multiplied by 2. So we get -64 plus 40. This value is negative. And this means that our height is decreasing. So basically we have verified that time of 54s is the point of maximum because the function was increasing and then started to decrease after. 5 words, right? We have identified the point of maximum. So now we can evaluate the maximum height H max. is found at 54, so we're basically substituting T equals 5/4s into our equation. We're going to get -16. Multiplied by 5 quarts squared. Plus 40 multiplied by 5 forts. Let's simplify -16 multiplied by 25 divided by 16. Plus 40 multiplied by 5 divided by 4. So for the first term, we got -25. For the second term, we get 40 divided by 4, which is 10. And send multiplied by 5 is 50, so we're adding 50. Now -25 + 50 gives us 25. The units are feet because that's height. And therefore, the correct answer to this problem corresponds to the answer choice B 25 ft. Thank you for watching.

Key Concepts

Quadratic Functions

Vertex of a Parabola

Maximizing Height

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