math Algebra 2 compettition

Question

A competition is held to see who can throw a baseball the highest from the front row of the upper deck seats at the stadium, 83 ft above field level. The winner throws the ball with an initial vertical velocity of 92 ft/sec and it lands on the infield grass.

 

Using the free-fall motion function, h(t) = -16t² + v₀ t + h₀ , where t is the time in seconds, h(t) is the height above the ground in ft, v₀ is the initial vertical velocity in ft/s, and h₀ is the initial height in ft, find out how many seconds it takes for the ball to hit the ground.

 

Type your answer into the box, rounded to the nearest hundredth.

Answer 1

h(t)=-16t^2+92t+83=0

complete the square

-16(t^2+23t/4 +23^2/64) + 83 +23^2/64 =0

-16(t-23/8)^2 + [(83)(23) +23(23)]/64 =0 in vertex form

-16(t-2 7/8)^2 + ... =0

time to reach max height = 2.875

time to reach the ground is a little over twice that, maybe about 6 or so seconds

-16t^2 +92t +83 =0

use the quadratic formula or factor if you can

-b/2a+(1/2a)sqr(b^2-4ac) where b=92, a =-16, c = 83

t=92/32 +(1/32)sqr(92^2+64(83)) = time to reach the ground

ignore the negative square root

t =23/8 +[sqr13776]/32 =about 2.875+3.668

= about 6.54 seconds to hit the ground