initial velocity

Question

If an archer shoots an arrow straight upward with an initial velocity of 128 ft/sec from a height of 6 ft, then its height above the ground in feet at time t in seconds is given by the function h(t)=-16t^2+128t+6. 

 

a. What is the maximum height reached by the arrow? 

b.How long does it take for the arrow to reach the ground?

 

Simplify your answer. 

 

help is greatly appreciated, thank you!

Answer 1

Hi Laura, starting with the function given, h(t) = -16t² + 128t + 6, we need to find the height at the top of the arrow's flight.

 

Think of the arrow's flight as an upside down parabola. The vertex of the parabola will represent the maximum hight the arrow will go. The vertex is given by: (h, k), where h = -b/2a and k can be found once we have a value for h.

 

h = -b/2a = -128/(2 * -16) = -128 / - 32 = 4

 

Substitute 4 for t in the given function will give a value for k.

 

h(4) = -16 * 4² + 128 * 4 + 6

 

       =  -256 + 512 + 6

 

       =  262

 

The vertex is: (4, 262), which means the arrow will reach it's highest point in 4 seconds at a height of 262 feet.

 

How long does it take the arrow to hit the ground. About 8 seconds, 4 seconds up and 4 seconds down.

 

To be more exact, use the quadratic formula to find the solution.

 

t = (-b ± √(128² - 4 * -16 * 6)) / (2 * -16)

 

t = (-128 ±√16768) / -32

 

t = (-128 ± 129.49) / -32

 

t₁ = -257.49 / -32 ≈ 8.05 seconds

 

t₂ = 1.49 / -32 ≈ -.047 seconds, this means the arrow hit the ground before it was shot and the solution can be discarded.

 

Final solution:

 

Part a:

 

The arrow reaches a peak of 262 feet.

 

Part b:

 

The arrow will hit the ground is approximately 8.05 seconds.

 

Questions?