Aball is tossed straight up with an initial velocity of 13.6 m/s. a) how high will it go? b) how long is the ball in the air?

A: 9.4248 m B: 2.773 s A ball tossed in to the air at 13.6m/s will experience a constant acceleration due to gravity of 9.81m/s^2. It will reach a 0 velocity at 13.6m/s^2 / 9.81m/s^2 = 1.386 s from the time it was thrown. It will fall to the ground another 1.386 seconds giving a total time in the air of 2*13.6m/s^2 / 9.81m/s^2 = 2.773 s To calculate the maximum height of the ball we use the fact that acceleration is constant. Between being thrown and reaching the maximum height the velocity constantly accelerates from 13.6m/s to 0 which gives an average velocity of 13.6m/s / 2 = 6.8m/s Now that we know average velocity and the time, the distance can be calculated. 1.386s * 6.8m/s = 9.4248m