Trajectory high point A stone is launched vertically upward fr...

Question

Trajectory high point A stone is launched vertically upward from a cliff 192 ft above the ground at a speed of 64 ft/s. Its height above the ground t seconds after the launch is given by s = -16t² + 64t + 192, for 0 ≤ t ≤ 6. When does the stone reach its maximum height?

Accepted Answer

Welcome back, everyone. At what time does the bowler reach its maximum height if it is thrown upward from the top of a 50 m high building with an initial speed of 20 m per second and its height above the ground is given by the equation S of T equals -10 T2 + 20 T + 50? We're given 4 answer choices A says T equals 0 seconds, B 1 2nd, C 2 seconds, and D 3 seconds. So essentially what we want to do is differentiate the position function as of T. We want to identify the derivatives of T. So we're going to take -10 multiplied by the derivative of T squared. We're going to add 20 multiplied by the derivative of T, and we're going to add the derivative of 50. So now applying the power rule, we're going to get the derivative of T squad, which is TT, that's -20 T when it is multiplied by -10. The derivative of T is 1, so we get Plus 20 and the rate of about 500 there is a constant. And now we want to set it to 0 to identify the critical points. So we're taking -20 T plus 20 equals 0. We can rearrange and we can say that 20 is equal to 20T and therefore dividing both sides by 20, our critical point T is equal to 1. What we now want to do is basically prove that this is a point of maximum. We can either understand that this is a parabola that opens down because it has a negative leading coefficient and its maximum is going to be at the vertex, or we can use the second derivative test and let's identify as double prime of T, which is the derivative of -20 T + 20. And once again, the derivative of T is 1, so we get -20 + 0 because the derivative of 20 is 0. And this is negative for any value of T, which means that the function is indeed concaved down for any T, and this proves that T equals 1 is a point of maximum. So the correct answer to this problem corresponds to the answer choice B. Thank you for watching.

Trajectory high point A stone is launched vertically upward from a cliff 192 ft above the ground at a speed of 64 ft/s. Its height above the ground t seconds after the launch is given by s = -16t² + 64t + 192, for 0 ≤ t ≤ 6. When does the stone reach its maximum height?

Key Concepts

Quadratic Functions

Vertex of a Parabola

Calculus and Optimization

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