Parametric equation problem?

• Apr 2nd 2011, 11:54 AM
homeylova223
Parametric equation problem?
The human cannonball is shot out of a cannon with an initial velocity of 70 mph 10 feet above the ground at an angle of 35 degrees.

What is the maximum range of the cannon?

I did this parametric equation
x=t v cos
y= t v sin-1/2gt^2+h

So I set y to zero and to find t and I got t= 3.14
Then I plug it in back to t in and got x=180.04
As I checked the answer so I am not sure what I may have done wrong?

These are the formulas I used
x=t 70 cos (35)
y= t 70 sin (35)-16t^2+10
• Apr 3rd 2011, 07:28 AM
Soroban
Hello, homeylova223!

Quote:

The human cannonball is shot out of a cannon with an initial velocity of 70 mph
10 feet above the ground at an angle of 35 degrees.

What is the maximum range of the cannon?

$\text{These are the formulas I used:}$

. . $x \:=\:(70\cos35)t$ . no

. . $y \:=\:10 + (70\sin35)t - 16t^2$ . no

$\text{I set }y\text{ to zero and to find }t\text{ and I got: } t= 3.14$ . no

"They" pulled a dirty trick on us!

The height is in feet . . . the gravitational constant is in ft/sec $^2.$

But the velocity is in miles per hour!

We find that 70 mi/hr is equal to $\frac{308}{3}$ ft/sec.

The equations are: . $\begin{Bmatrix}x &=& \left(\frac{308}{3}\cos35\right)t \qquad\qquad\quad \\ \\[-3mm]
y &=& 10 + \left(\frac{308}{3}\sin35\right)t - 16t^2 \end{Bmatrix}$

Now we can proceed . . .

• Apr 3rd 2011, 07:30 AM
HallsofIvy
Excellent, Soroban, I missed that completely!
• Apr 3rd 2011, 07:40 AM
homeylova223
Thanks Soroban.
• Apr 3rd 2011, 09:22 AM
homeylova223
I have one more question
I set y to zero ended up getting t=4.305
then plugged that back into x equation to get x=362.086 ft
While apparently the correct answer is 323.2 ft
• Apr 3rd 2011, 09:51 AM
skeeter
Quote:

Originally Posted by homeylova223
I have one more question
I set y to zero ended up getting t=4.305
then plugged that back into x equation to get x=362.086 ft
While apparently the correct answer is 323.2 ft

$0 = 10 + \dfrac{308}{3} \sin(35) \cdot t - 16t^2$

$a = -16$ , $b = \dfrac{308}{3} \sin(35)$ , $c = 10$

$t \approx 3.84$ sec
• Apr 3rd 2011, 10:35 AM
homeylova223
I am slightly confused because I did this

0=10+308/3 sin 35-16t^2
0=10+58.887t-16t^2 factor t
0=68.887-16t
16t=68.887
t=4.305

What error could have been made?
• Apr 3rd 2011, 11:02 AM
skeeter
Quote:

Originally Posted by homeylova223
I am slightly confused because I did this

0=10+308/3 sin 35-16t^2
0=10+58.887t-16t^2 factor t
0=68.887-16t
16t=68.887
t=4.305

What error could have been made?

$10 + 58.887t \ne 68.887 t$