An object is launched straight upward from a platform. Its height above ground is h(t) = h0+ v0t ? 16.1t2 where h is in feet and t is in seconds. The graph of h is shown below. Find both unknown constants, accurate to 1 decimal place. Solution h(t) = h0 + v0t - 16.1t^2 As can be seen, the maximum height is reached at t = 2.5 : So, -b/(2a) = 2.5 -v0 / (2*-16.1) = 2.5 v0 / (32.2) = 2.5 v0 = 2.5 * 32.2 v0 = 80.5 ---> Answer for v0 When t = 1 , h = 90 90 = h0 + 80.5*1 - 16.1*1^2 h0 = 90 - 80.5 + 16.1 h0 = 25.6 So, the constants are : v0 = 80.5 AND h0 = 25.6.