Using a Vertical Motion Model

Question

A baseball player hits a foul ball straight up in the air from a height of four feet off the ground. The initial velocity as the ball comes off the bat is 130 feet per second. The motion of ball can be modeled by y=-16² + vt + h, where t is the time in seconds, v is the initial velocity in feet per second, and h is the height off the ground in feet. 

write a quadratic function that models the height of the ball in terms of time.

what is the height of the ball after 1 second? What is the height of the ball after 2 seconds? What is the height of the ball after 10 seconds? Show all of your work.

do all of your answers to question to make sense? use complete sentences in your answer.

write an equation that you can use to determine when the ball hits the ground. Then solve the equation. Show all your work.

when does the ball hit the ground? Use complete sentence in your answer.

when is the ball at its highest point. Show your work

what is the greatest height of the ball? show your work.

Answer 1

h(t) = h0 + v0*t + (1/2)a*t^2

 

h0 = 4 ft, v0 = 130 ft/s, a = -32 ft/s^2

 

h(t) = 4 + 130t - 16t^2, with t in seconds, h in ft

 

h(1) = 4 + 130(1) - 16(1)^2 = 118 ft

 

h(2) = 4 + 130(2) - 16(2)^2 = 200 ft

 

h(10) = 4 + 130(10) - 16(10)^2 = -296 ft

 

(This last answer makes sense only if the ground was not present to stop the ball from falling further below zero height.  The other answers make sense.)

 

Ball reaches ground when h = 0

 

h = 0 = -16t^2 + 130t + 4

 

Solve using quadratic formula:

 

A = -16

B = 130

C = 4

 

t = [-B ± √(B^2 - 4AC)] / (2A)

 

B^2 - 4AC = 130^2 - 4(-16)(4) = 17156

 

√(B^2 - 4AC) = √(17156) = ±130.98

 

2A = 2(-16) = -32

 

t = (-130 ± 130.98) / (-32)

 

t > 0 so use the negative square root:

 

t = (-130 - 130.98) / (-32) s = 8.16 s, which is time at which ball reaches ground

 

Ball is at its highest point when v = 0.

 

v(t) = v0 + at = 130 - 32t = 0

 

t = 130/32 s = 4.06 s

 

Ball is at its highest when t = 4.06 s.

 

h_max = h(4.06) = 4 + 130(4.06) - 16(4.06)^2 = 268.06 ft