Question 723515
An arrow shot vertically into the air from ground level with a crossbow reaches a maximum height of 324 feet after 4.5 seconds of flight. Let the quadratic function d(t) represent the distance above ground (in feet) t seconds after the arrow is released. (If air resistance is neglected, a quadratic model provides a good approximation for the flight of a projectile.)
:
(A) Find d(t)
Find the initial velocity (v) of the arrow using the form -16t^2 + vt = d(t)
t=4.5, d(t)= 324
-16(4.5^2) + 4.5v = 324
-16(20.25) + 4.5v = 324
-324 + 4.5v = 324
4.5v = 324 + 324
4.5v = 648
v = 648/4.5
v = 144 ft/sec is the initial upward velocity of the arrow
the equation
d(t) = -16t^2 + 144t
:
(B) At what times (to two decimal places) will the arrow be 250 feet above the ground?
-16t^2 + 144t = 250
-16t^2 + 144t - 250 = 0
Solve for t using the quadratic formula, 
I got t = 2.35 and 6.65 sec at 250 ft
:
Graphically, green line is 250 ft, blue line is 324 ft
{{{ graph( 300, 200, -5, 15, -100, 400, -16x^2+144x, 250, 324) }}}