SOLUTION: If a ball is thrown straight up from the top of a building that is 407 feet high, the position in feet above the ground is given by the function s(t) = -16t^2 + 75t + 407 where t

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: If a ball is thrown straight up from the top of a building that is 407 feet high, the position in feet above the ground is given by the function s(t) = -16t^2 + 75t + 407 where t       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1115150: If a ball is thrown straight up from the top of a building that is 407 feet high, the position in feet
above the ground is given by the function s(t) = -16t^2 + 75t + 407 where t is the number of
seconds elapsed.
a. How high is the projectile after 3 seconds?
b. How long will it take for the ball to reach a height of 450 feet above the ground?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
If a ball is thrown straight up from the top of a building that is 407 feet high, the position in feet above the ground is given by the function
s(t) = -16t^2 + 75t + 407 where t is the number of
seconds elapsed.
a. How high is the projectile after 3 seconds?
s(3) = -16*3^2+75*3+407 = 488 ft
----------------------------------------
b. How long will it take for the ball to reach a height of 450 feet above the ground?
Solve:: -16t^2+75t+407 = 488
-16t^2+75t - 81 = 0
t = 1.6875 seconds
-----
Cheers,
Stan H.
-----------