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.
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
t=15/9.8 for max height.
When position of y = 0 the ball reaches its distance and final duration so:
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.
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.