Calculus based kinematics (two dimensions) question

[RJLiberator]

Homework Statement

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.

Homework Equations

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.

 

[Brian T]

Everything looks right but you really calculated the max height time twice. Once by setting the velocity equation r'_y = 0, you solved and got t = 15/9.8

The other by solving for the full time t = 15/4.9 then dividing by 2.

These are the same times so you only had to go wih one method. If you want simpler calculations, leave the 9.8 as g so you only have to multiply it through at the way end (or leave it in terms of g if your teacher doesn't mind)

 

[RJLiberator]

Damn, so since everything is correct, I have to assume they either want it simplified via use of a calculator (forbidden generally in this course) or unsimplified and a very whacky answer.

I guess it is good that I understand the problem :P.

I will have to talk to my instructor about this.

 

[Brian T]

15*15/98-9.8/2*(15/9.8)^2=11.48m

Damn, so since everything is correct, I have to assume they either want it simplified via use of a calculator (forbidden generally in this course) or unsimplified and a very whacky answer.

I guess it is good that I understand the problem :P.

I will have to talk to my instructor about this.

You could go a bit further in terms of simplifcation w/o calculator:
Using y = v_ot + (1/2)at²
$y = 15(\frac{15}{9.8}) - \frac{9.8}{2}(\frac{15}{9.8})^2$

Cancel out the 9.8's on the right term, collect the squared terms:
$y = (\frac{15^2}{9.8}) - \frac{1}{2}(\frac{15^2}{9.8})$

Factor our (15^2)/9.8 which gives us:
$y = \frac{225}{9.8}(1 - \frac{1}{2})$
$y = \frac{225}{19.6}$

And then a little long division to finish it off? :D