Max height expressed with v, theta, and g

[uber_dzl]

[SOLVED] max height expressed with v, theta, and g

Homework Statement

what is the max height of a projectile with velocity v at an angle theta using the conservation of energy? express in terms of v, theta, and g

Homework Equations

1/2*m*v^2=mgh

The Attempt at a Solution

i tried and got (v^2*sin(theta))/2g (wrong)

 

[blochwave]

You've got the right idea. In fact I think you're close. Actually type out your work and show us, see if you spot the mistake

 

[uber_dzl]

((1/2)m(v^2)*sin(theta))=mgh...then ((1/2)m(v^2)*sin(theta))/mg=mgh/mg...then the m cancels leaving me with (v^2*sin(theta))/2g=h

 

[uber_dzl]

is my issue the placement of sin(theta) in the equation before i solve for h?

 

[uber_dzl]

wow! what a tedious mistake...it was the placement of the exponent and parenthesis that was wrong...


thanx

 

[blochwave]

Well my issue was you start here

1/2*m*v^2=mgh

It's the VERTICAL velocity you care about, it's 1/2*m*Vy^2=mgh

Vy=Vsin(theta) so you get 1/2*m*V^2sin^2(theta)