A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 39.2 m ...

A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 39.2 m above the ground. The projectile strikes the ground with a speed of 1.3v0. Find v0. (Ignore any effects due to air resistance.)

v0 = m/s

.5mv^2 + m*9.8*39.2 = .5m(1.3v)^2
.5v^2 + 9.8*39.2 = .5(1.3v)^2
v^2 + 2*9.8*39.2 = 1.69v^2
.69v^2 - 2*9.8*39.2 = 0

v = 33.2 m /s

Here's a good formula to remember for a projectile:

If a cannon is fired at an inclination angle of alpha with the horizontal plane and with an initial velocity of Vo, then:
Vx = Vocos alpha and Vy = Vosin alpha
Then position of projectile at any time t is given by the parametric equations:
x = (Vocos alpha)t and y = -.5gt^2 + (Vosin alpha)t

Eliminating t from these two equations gives:

y = gx^2/(2Vo^2cos^2 alpha) + xtan alpha

Vf² = Vo² + 2gh → (1.3Vo)² - Vo² = 2gh = Vo²[1.3² - 1.0] →

Vo = √[2gh/0.69] = 33.37 m/s