The path of a large arrow fired from a catapult can be modeled by y= -0.0044x^2+1.68x, where x is the distance the arrow traveled (in yards) and y is the height of the arrow (yards). Give the height of the castle wall, find the safest distance from the wall to launch an arrow over the wall.
1. the height of the wall is 120 yards
2. the height of the wall is 100 feet.

Step-by-step explanation:

Wwe can solve this in either of two ways:  Differentiation or graphing.

Differentiation

y= -0.0044x^2+1.68x

y' = -0.0088x + 1.68

y' is an equation that tells us the slope of the line for any point x.  We want to know when the arrow begins it's descent.  That is the point of maximum height, and the slope of the line through that point will be zero.  Set y' = 0 and solve for x:

0 = -0.0088x + 1.68

0.0088x = 1.68

x = 191

The maximum height is when the arrow has travelled 191 feet.  Now put 191 feet into the equation to find the maximum height:

y= -0.0044x^2+1.68x

y(191) = feet

The maximum height is 160 feet

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Graphing

See attachment.

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v