Math problem

Question

 define variables, write an equation, and then use the equation in the solving process.

projectile is thrown upward so that its distance, in feet, above the ground after t seconds is h= -10t² +260t

What is its maximum height?

Answer 1

The equation of motion is unusual: it suggests that the acceleration due to gravity is 20 ft/sec2; the usual value would be 32 ft/sec2.

Taking the equation as given:

If you have calculus, the derivative is -20t + 260 = 0 at the maximum height, i.e. at t =13 sec.

If you do not have calculus, you need to know that the vertex of a parabola occurs at -b/2a where the parabola is :
at² + bt + c, which for this problem is -260/-20 = 13 sec.

Substitute 13 into the original equation to get the maximum height of 1690 feet.

Answer 2

h = -10t² + 260t

Since it is a parabola, find the vertex. The y-value will be the maximum height

t = -b/2a

t = -260/(-20)

t = 13 seconds

 

h = -10(13)² + 260(13)

h = -1690 + 3380

h = 1690 feet