1. The problem statement, all variables and given/known data A toy rocket, launched from the ground, rises vertically with an acceleration of 33 m /s^2 for 13 s until its motor stops The acceleration of gravity = 9.8 m/s^2 Disregarding any air resistance, what maximum height above the ground will the rocket achieve? Answer in units of km. G = -9.8 m/s^2 A = 33 m/s^2 with T = 13 seconds Unknown x - xi = ? 2. Relevant equations x - xi = vi * t + 1/2 A T^2 V = A * T V = Vi - gt 3. The attempt at a solution Ok, so first I broke this problem up into two pieces, I checked the distance it traveled while accelerating by using the first equation I provided... the Vi * t = 0 since my Vi was 0 and the rest I got --- 1960.4 m then I saw what velocity it was going at that time and I got 301.6 m/s So then I needed to find the distance it traveled from 301.6 m/s to 0 m/s so I used the third equation and set V = 0 to find T which equals 30.7755102 seconds. Then I reused the 1st equation to find the distance of this part By plugging g in for my acceleration component of course, I got 4,640.946939 m + my first part 6,601.346939 m / 1000 = 6.601.34 km However this answer came up as wrong, ( I still have 6 tries left!) I can't see where I went wrong? I believe it's in the 2nd part finding how much distance the velocity goes until it reaches 0. Thanks for your help!