Question

A ball 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?

Answer 1

For this case we have the following equation:
 h (t) = -16t2 + 40t + 1.5
 To hit the ground we have:
 -16t2 + 40t + 1.5 = 0
 Solving the polynomial we have:
 t1 = -0.04
 t2 = 2.54
 Taking the positive root (for being time) we have:
 t = 2.54 s
 For the maximum height we have:
 We derive the equation:
 h '(t) = -32t + 40
 We equal zero and clear t:
 t = 40/32
 t = 1.25
 We evaluated t = 1.25 in the function:
 h (1.25) = -16 * (1.25) ^ 2 + 40 * (1.25) + 1.5
 h (1.25) = 26.5 feet
 Answer:
 the ball hit the ground at:
 t = 2.54 s
 The maximum height of the ball is:
 h (1.25) = 26.5 feet

Answer 2

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

• The time after which the ball would hit the ground is approximately 2.54 seconds

• The maximum height of the ball is 26.5 feet

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;

$x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

Therefore;

$t = \dfrac{-40\pm \sqrt{40^{2}-4\times (-16)\times 1.5}}{2\times (-16)}= \dfrac{-10\pm\sqrt{106}}{-8}$

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;

$x = -\dfrac{b}{2 \cdot a}$

$t = -\dfrac{40}{2 \times (-16)} = 1.25$

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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