Theory and Examples

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Question

Theory and Examples

Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀, g > 0, with s in meters and t in seconds. Find the body’s maximum height.

Accepted Answer

Welcome back, everyone. A ball is thrown vertically upward from the ground. The height of the ball in meters, 2 seconds after it has been thrown is given by Y equals -10 GT 2 plus UT. Given that G equals 10 m per second squared and U equals 50 m per second, calculate the maximum height reached by the ball. A says 6.25 m, B 12.50 m, C, 18.75 m. In the 25.00 m. We're given the position function, and if we recognize, well, this is a quadratic polynomial. Because we have a negative leading coefficient, we can visualize the trajectory as a parabola that opens down. We're going to reach the maximum height at the vertex. So we want to identify the time value at the vertex, and let's call it T1. What we want to do is simply identify the velocity function, which is the derivative of the position function. So let's differentiate negative 10 GT2 plus UT. Our independent variable is T, so we can apply the power rule which gives us negative 10G, that's our constant, multiplied by the derivative of T2, which is 2T. And then we're adding U, which is a constant multiplied by the derivative of T, which is 1. So our velocity function is -20 GT plus U. At the highest point, our velocity becomes 0. So what we want to do is size 0 equals -20 GT1 plus U. So, in this context, we're saying that at T1 velocity is 0. So let's solve for T1. If we rearrange the equation, we can set. 20 GT 1 equals U. And therefore T1 Is equal to U divided by 20 G. Now that we have our time value, we can substitute it into the position function and evaluate the maximum height H max. Which is found by taking time value of you divided by 20 G. So for every time value that we have, we're going to substitute you divided by 20 G, which gives us negative 0 G multiplied by U2 divided by 400g squad. Here we are squaring U divided by 20 G. And we're adding U multiplied by T, which is U multiplied by U divided by 20 G. Now, let's simplify the result. We can cancel out G and G squad for the first term. This essentially gives us Negative 10 U 2 divided by. 400G. Plus you squared. Divided by 20 G. And now what we can see is simplifying the common denominator. Which is going to be 400G. So we are taking 20 U 2 essentially for the second fraction, we need to multiply both sides by 20. So this gives us 20 U squad, and we are subtracting. 10 you square. Which gives us our final expression, that's 10 squared. Divided by 400G. Or simply U squared divided by 40 G. And now we know you, we know G, so we can perform substitution. This gives us 50. meters per second squared. Divided by 40. Which is multiplied by G and that's 10 m per second squared. When we perform the calculation, we're going to get 6.25 m. Which corresponds to the answer to a. Thank you for watching.

Theory and Examples

Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀, g > 0, with s in meters and t in seconds. Find the body’s maximum height.

Key Concepts

Quadratic Functions

Vertex of a Parabola

Derivative and Critical Points

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