A baton twirler tosses a baton into the air. the baton leaves the twirlers hand 6 feet above the ground and has an initial velocity of 45 feet per second. the twirler catches the baton when it falls back to a height of 5 feet. For how long is the baton in the air?

Let's find out the ball's acceleration V^2 = U^2 + 2as Where the distance S = 6 feet and U = 45. Suppose he just took off so V = 0 and we are left with U^2 = 2as (45) ^2 = 2a (6) a = 2025/12 = 168.75 ft/s^2 Also S = ut + 1/2 at^2 Here S = 5ft and U = 0; a = 168.75. So we have 5 = (1/2) * 168.75 * t^2 So we have 5 * 2 = 168.75 t^2 T^2 = âš 10 / 168.75 = 0.00596 = âš0.006 T = 0.0077s