Thread: 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

Hello, homeylova223!

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?




. . . no

. . . no

. no

"They" pulled a dirty trick on us!

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

But the velocity is in miles per hour!


We find that 70 mi/hr is equal to ft/sec.


The equations are: .

Now we can proceed . . .

Excellent, Soroban, I missed that completely!

Thanks Soroban.

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

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

your time is incorrect.



, ,

sec

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?

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?

... the quadratic will not factor. use the quadratic formula.