Like before we can get the position function, x(t), by integrating the acceleration, which in this case is gravity g = 9.81m/s². We will assign the downward direction as negative x in this problem and let the ground be x=0. Integrating once gives velocity: V(t) = -9.81t +17.
Integrating this to get position x(t) gives
x(t) = (-9.81/2)t² +17t = 4.905t² +17t
We need to solve this quadratic equation for the time t when the position x(t)=0 because this is when the ball hits the ground. Using the quadratic formula gives
t= 0 and 3.466
All quadratic formulas have two roots (answers) Here the 0 root just means the ball was at the ground at t=0 (which it was because that's where it was thrown from) The other root is of interest because that's when it lands back on the ground and is our answer to part a.
t = 3.466s
To get part b we only now need to plug in half of our time to the position function. Half because it went up for half the time and down for half. This gives
x(t) = 4.905(3.466/2)² +17(3.466/2) = 44.19m
part c is asking how long it takes to get to the highest point. We already found that out in part b. It is half the time from part a: t = 1.733s
