Throwing a stone Suppose a stone is thrown vertically upward f...

Question

Throwing a stone Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t²+32t+48.
b. When does the stone reach its highest point?

b. When does the stone reach its highest point?

Accepted Answer

Welcome back, everyone. A diver jumps off a diving board 10 ft above the pool with an initial upward velocity of 16 ft per second. The height and feet of the diver above the water 10 seconds after the jump is given by S of T equals -16 T2 + 16 T + 10. When does the diver reach the maximum height? A 1 2nd, B 2 seconds. C 0.5 seconds, and D 3 seconds. So now, before we begin, we have to visualize this as a function. If we plot a coordinate plane. Well, we can understand that this is a parabola. That opens down, right? We essentially have our starting point at C over 0, that's 10. Since the para opens down, we start increasing, then we start to decrease until we reach. The minimum height, right, essentially 0 at some point. Now, the maximum height is basically the vertex of this parabola by definition. And what we can notice is that the maximum height. is achieved when the tangent line is horizontal, right? So, essentially we are going to differentiate the function. We're going to find S T, which is the velocity function because this is the rate of change of height. Let's identify the derivative. We want to differentiate -16 T2 + 16 T + 10. We can apply the power rule, which gives us -302 T + 16 + 0 because we have a constant. And now we're setting the velocity function. Which is basically as prime of T. Equal to 0 because at that instant. The tangent line is horizontal, right? And it corresponds to the maximum height. So -302 T + 16. Is equal to 0. We can then rearrange and say that 16 is equal to 32 T. and solving for T, we take 16 and divide by 32, which gives us 1/2 or 0.5. Now, of course, we can, we can check the sign of the derivative. Notice that if T is less than 0.5. Then V of T, let's suppose that T is 0, right? We get positive V of T because if T equals 0, then V of T is 16 or specifically B of 0, that's what we're checking and that's positive. If time is greater than 0.5, let's post 1. Then we get -302 + 16, which is -16, and that's negative. So velocity keeps increasing or actual velocity is positive, and then it becomes negative after 0.5. What does that indicate? Well, it indicates. The point of extrema and specifically it is the point of maximum because the derivative changes its sign from positive to negative, meaning this is where we have the maximum height and we have proved without a sketch that time equals 0.5 seconds is the point of maximum height. So the correct answer to this problem. Corresponds to the answer choice C 0.5 seconds. Thank you for watching.

Key Concepts

Quadratic Functions

Vertex of a Parabola

Maximizing Functions

Watch next