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

The time after which the ball would hit the ground is approximately 2.54 secondsThe 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;

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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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
 
 the ball hit the ground at:
 t = 2.54 s
 The maximum height of the ball is:
 h (1.25) = 26.5 feet

Idon’t know actually i’m in 5th grade and don’t know

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