**1. The problem statement, all variables and given/known data**

A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 38.0 m above the ground. The projectile strikes the ground with a speed of 1.3v0. Find v0. (Ignore any effects due to air resistance.)

**2. Relevant equations**

Time= 0=v_{0}T-½gT^{2} T<0

t= (v_{Oy}+√(v_{Oy}^{2}-2gy))/g

distance/range x=x_{0}+V_{0x}t

x/y velocity- v_{0x}=Vcos(θ); v_{0y}=vsin(θ)

**3. The attempt at a solution**

This is a practice problem (with different values than original), and the provided answer is **32.9m/s**

i set the origin at the cannon

Since the velocity magnitude = √(Vcos(θ)^{2}+Vsin(θ)^{2})

I tried variations along the line of V_{0}cos(30)^{2}+V_{0}sin(30)^{2}= 1.3 V_{0}^{2}

in order to split the velocity into xy components.

Conceptually i understand it takes him -38m to reach 1.3 times his original velocity, but in trying to plug the provided data into the equations i end up with too many variables.

Any and all help very appreciated, thank you.

http://ift.tt/1iprbj1