SOLUTION: For a project I had to show an application of quadratics using the H(t)= gt^(2) +vt+h equation and projectile motion. I built a trebuchet and launched a softball. It traveled a d

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: For a project I had to show an application of quadratics using the H(t)= gt^(2) +vt+h equation and projectile motion. I built a trebuchet and launched a softball. It traveled a d      Log On


   

Question 704374: For a project I had to show an application of quadratics using the H(t)= gt^(2) +vt+h equation and projectile motion. I built a trebuchet and launched a softball. It traveled a distance of 83 1/6 feet in 3.1 seconds. Its intial height was 7/12 feet. I have added my data and my equation is h(t)= -16t^(2) + 2495/93t +7/12. I'm confused on how to simplify from here. Could you please explain it to me? Thank you
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
For a project I had to show an application of quadratics using the H(t)= gt^(2) +vt+h equation and projectile motion. I built a trebuchet and launched a softball. It traveled a distance of 83 1/6 feet in 3.1 seconds. Its intial height was 7/12 feet. I have added my data and my equation is h(t)= -16t^(2) + 2495/93t +7/12
:
-16t^2 + vt + 7/12 = 0; (h=0 when it hits the ground)
It took 3.1 seconds, (t=3.1), find the velocity
-16(3.1^2) + 3.1v + 7/12 = 0
-153.76 + 3.1v + 7/12
3.1v = +153.76 - 7/12
3.5v = 153.1767
v = 153.1767/3.1
v = 49.4 ft/sec is the velocity
the equation for the device then
h(t) = -16t^2 + 49.4t + 7/12
:
If you graph this
+graph%28+300%2C+200%2C+-2%2C+5%2C+-10%2C+50%2C+-16x%5E2%2B49.4x%2B%287%2F12%29%29+
Note it strikes the ground in 3.1 sec, reaches a max height of 38.7 ft (the vertex)