Projectile motion of a cannonball problem

[sweet_girl123]

a cannonball is fired with initial velocity v0 at an angle 30 above the horizontal from the height of 40m above the ground. the projection strikes the ground with a speed of 1.2*v0. find v0

 

[sweet_girl123]

I tried by getting the time first
y = 1/2 gt^2

but I am lost after that can anyone please help me out...

 

[VulcanWong]

Energy consideration maybe another(also faster i think=)) method to get the answer.
Try to think about the whole system energy(involve both KE and PE).Using the motion equation(Your Method), you could also get your answer eventually.
Think about the VERTICAL motion(Vo sin 30), you will also get the answer by setting up a motion equation

 

[I like Serena]

Hi sweet_girl123!

Easiest way to do this is with energy conservation (as VulcanWong suggested).

The formula for that is:
$$g y_i + {1 \over 2} v_i^2 = g y_f + {1 \over 2} v_f^2$$
with:
$y_i, y_f$ the initial and final heights
$v_i, v_f$ the initial and final (total) speeds
$g = 9.8 {m \over s^2}$ the acceleration of gravity

Can you fill in the numbers?