Aball is shot from a cannon into the air with an upward velocity of 40 ft/sec. the equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t2 + 40ft + 1.5. • about how many second after launch would the ball hit the ground? • what is the maximum height of the ball?

The shape of the graph of the given function is a parabola with the maximum point located at the vertex

Reasons:

The upward velocity of the ball, u = 40 ft./s

The height (h) of the ball at any time (t) is, h(t) = -16·t² + 40·t + 1.5

Required:

(a) The time after launch after which the ball will hit the ground

Solution:

The height of the ball when the ball hits the ground is zero, therefore

At the ground, h(t) = 0 = -16·t² + 40·t + 1.5

From the quadratic formula, we have;

Therefore;

t ≈ 2.537 or t ≈ - 0.037

Therefore

The number of seconds after launch after which the ball would hit the ground, t ≈ 2.54 seconds

The maximum height of the ball, is given from the x-value of a quadratic

function f(x) = a·x² + b·x + c, at the maximum height as follows;

Therefore;

h(1.25)  = -16×1.25² + 40×1.25 + 1.25 = 26.5

The maximum height of the ball is 26.5 feet

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