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Question

Tyler threw the ball in to the air. It landed 2.6 seconds later. He released the ball at 4.5 feet. Figure out a function to model the ball. Only algebraic solutions accepted.

Answer 1

The initial height is 4.5 feet.  This is the point (0, 4.5).

The x-intercept is (2.6, 0).  This is the landing point.

 

The equation is a parabolic behavior. Because it goes up then down.

 

y = -16t² + bt + 4.5

 

is what the equation looks like so far.  The leading coefficient is the standard acceleration due to gravity.

 

 

Now, we set y=0 and t=2.6 to solve for b.

 

0 = (-16)(2.6)² + (2.6)b + 4.5

 

0 = -108.16 + 2.6b + 4.5

 

103.66 = 2.6b

 

39.87 = b

 

The function of the trajectory is

 

y = -16t² + 39.87t + 4.5

 

for     t ≥ 0

Answer 2

The ball will follow a parabolic path.

Height as a function of time will be:

h(t) = (-1/2)gt² + V₀t + h₀

g = acceleration of gravity = 32 ft/sec

t = time in seconds

v₀ = initial velocity

h₀ = initial height = 4.5 feet

 

h(t) = -16t² + v₀t + 4.5

 

h(2.5) = 0

Use this to solve for initial velocity:

 

-16(2.5)² + v₀t + 4.5 = 0

-100 + v₀(2.5) + 4.5 = 0

2.5v₀ - 95.5 = 0

2.5v₀ = 95.5

v₀ = 95.5/2.5

v₀ = 38.2 ft/sec initial velocity of the ball

 

h(t) = -16t² + 38.2t + 4.5

 

Check:

h(2.5) = -100 + 95.5 + 4.5 = 0

This checks.