1. The problem statement, all variables and given/known data A soccer ball is kicked from the ground at an angle of 30◦ with speed 30m/s. Determine the duration of the flight, range and maximal height of the trajectory. 2. Relevant equations 3. The attempt at a solution 1. Break down the velocity vector into x and y components: 30cos(30) = 15sqrt(3) m/s 30sin(30) = 15m/s r(t) = (0,0)+t(15sqrt(3), 15)+(t^2)/2(0,-9.80) This equation represents the position function. If I take the derivative to get the velocity function, it becomes r'(t) = (15sqrt(3), 15-9.8t) When velocity in the y direction is 0, then the ball is at max height so 15-9.8t=0 t=15/9.8 for max height. When position of y = 0 the ball reaches its distance and final duration so: 15t-4.9t^2=0 t=0 or t= 15/4.9 clearly it is not t=0. since we have time of duration: t*velocity in x = range 15/4.9*15sqrt(3) = 79.5329 m max height is found by dividing the time of max range by 2 so thew time of max height = 15/9.8 and then using that as t in the r(t) position function and finding the y-component. The y-component comes out to be 11.48m. Seems reasonable. 15*15/98-9.8/2*(15/9.8)^2=11.48m Am I doing this right? My problem is that our calculus class requires no calculators. To me, it seems like this problem is begging the use of calculators unless I just don't simplify things (which would be fine to ). It just seems like I am missing something that'd make the calculations easier. Thanks.