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=v0T-½gT2 T<0
x/y velocity- v0x=Vcos(θ); v0y=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 V0cos(30)2+V0sin(30)2= 1.3 V02
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.