Finding maximum height

Question

You and a friend are hiking in the mountains. You want to climb to a ledge that is 20 ft above you. The height of the grappling hook you throw is given by h(t)=- 16t ^2-32t+5. What is the maximum height of the grappling hook? Can you throw it high enough to reach the ledge? 

Answer 1

Because you've identified this as a quadratic equation problem, I'll assume you are in algebra or algebra 2 (or just doing pre-calculus physics). However, this is a simple calculus problem, and your initial formula is in the form of simple velocity problems for objects affected by gravity... but there is a mistake.

 

Such formulas typically follow the form:

 

y = y₀ + v₀ x t + t² x g / 2

 

However, your formula indicates a negative initial velocity (-32t as the middle, first order term). This suggests that the hook was thrown downward at an initial velocity of 32f/s. In other words, your formula is messed up (and if solved as is, would result in an imaginary/nonsense answer).

 

For something thrown upward, that term should be positive. Correcting your formula gives us:

 

y = -16t² + 32t + 5

 

The first derivative of this formula (dy/dt) gives us:

 

dy/dt = -32t + 32

 

This equals the velocity of the hook at any moment. The highest point of the throw will be where the velocity equals 0.

 

v = -32t + 32 = 0

t = 1

 

The maximum height of the throw is achieved at one second. Plug that back into your original formula to solve for the actual height of your throw.

Answer 2

This is a physics problem.

The real and correct formula is: h(t) = (1/2)(-32 ft/s²)·(t²) +(v₀)·(t) + h₀

_{But math textbooks like to ignore teaching the real science and explaining the formula,}

_{and use simplified and sometime not truly correct equations.}

If you want to solve the science problem like a pure math problem -- well. okay.

_{But, Don't forget your units! The solutions is not correct without units.}

For a polynomial of form Ax^2 + Bx + C = 0

• A tells the end behavior.

Negative says it is down, the end behavior is down on the right.

• It is an even power (x²), so both the beginning and ending behavior are the same.

SO - It is a quadratic and a U shape opening down.

• The axis of symmetry is given by x = -b / 2a

Find -b/2a

That is the X value for the vertex of a parabola.

Plug that value into the formula for h(t) and find h(-b/2a).

That is the Y value of the vertex or in this case the max height.

Is it more than 20 feet?