Find the Maximum height reached

Question

A stode is thrown upward so that any time x (in second) after throwing, the height of the stone is y=100+10x-5x^2 (in meter) . Find the maximum height reached?

Answer 1

Very nicely done Jaques. Another way to think about this problem is by remembering the relationship between position, and velocity. Let me re-write the equation given for clarity by substituting values height for y and time for x...

y=100+10x-5x^2 (meters)

height = 100+10*time - 5*time^2

This problem's equation is really a model of position which is dependent upon time. If we take the first derivative of this equation we will end up with a model of the change in position dependent upon the change of time.

change of height = 10 - 10*time

We can solve this model for when it is equal to zero to determine when (what time) the change of height is equal to zero (aka when the stone has reached its peak height e.g. no longer moving up or down).

0 = 10 - 10*time

Solving this equation we come to the conclusion that the stone reaches its peak height at time = 1s.

We then substitute that value back into the original second-degree polynomial equation given in the prompt and find our answer to be

height = 100 + 10*(1) - 5*(1)^2 = 105 meters!

Answer 2

The equation is a downward parabola, so the max is at the vertex. y = -ax²+bx+c ymax occurs at x = b/2a = 1. Plug this into equation and find ymax.

In physics, you know that the max height will occur at t = v₀/a in the equation y = y₀+v₀t - 1/2at² here v₀ is 10 m/s and a = 10 m/s² Using v = v₀-gt, you are solving for v=0 at apex.