Calculating Maximum Range of Projectile Fired at 210 m/s on Horizontal Ground

[pianogirl]

Homework Statement

Shells are fired from a gun at 210 m/s. What is the maximum range of the shells on horizontal ground?

Homework Equations

Suvat equations?

The Attempt at a Solution

x=210cos(theta)t
y=210sin(theta)t - 4.9t^2
Not sure what to do after that.
Help would be appreciated.

 

[tiny-tim]

Hi pianogirl!

(have a theta: θ and try using the X² tag just above the Reply box )

Shells are fired from a gun at 210 m/s. What is the maximum range of the shells on horizontal ground?
…
x=210cos(theta)t
y=210sin(theta)t - 4.9t^2
Not sure what to do after that.
Help would be appreciated.

Hint: put y = 0.

 

[denverdoc]

There is a more direct way to get time IMO than your eqn.

Consider only the ascent phase.

What happens when we reach the peak of flight to component of the velocity?

Doing so will give you an expression where t=(210 Cos(theta))/g

The descent will take exactly as long so the total time is 2t.

Now substitute for this in your above eqn. You should get something times sin(theta)cos(theta)

There is a trig identity: sin (2theta)= 2 sin(theta) cos (theta)

Simplify your eqn using this. By considering what angle sin is at its maximum should be the answer.

EDIT: sorry didn't see your post Tiny-Tim

 

[pianogirl]

But I still have two unknowns...? The range and the angle... or did I do something wrong?

 

[tiny-tim]

But I still have two unknowns...? The range and the angle...

But you have two equations, so that's ok

EDIT: he he … beat denverdoc again!

 

[denverdoc]

Nope you should have two unknowns, but ask yourself what angle will make y maximal?

In other words sin (a) = max at 90 degrees so sin(2a)= max at what angle?