Considering ONLY the y-direction (up/down), the velocity in the y direction (v_{y}) is calculated using sin(30°) = opposite/hypotenuse = v_{y}/70 therefore v_{y} = 70sin(30°) = 35 ft/s.

The equation that relates initial velocity, final velocity, time, and acceleration is v_{yf} = v_{yi} + at where "a" in this case is the acceleration of gravity (or "g") which is -32.17 ft/s^{2}, v_{yi} is again the initial velocity in the y-direction, "t" is time, and v_{yf} is the final velocity in the y direction which will be zero since the ball goes up until it instantaneously stops to reverse direction and come back down. So, considering half of the ball's trip (from the ground up to the peak of it's travel in the y-direction), we can use the equation to solve for time:

v_{yf} = v_{yi} + at

v_{yf} - v_{yi} = at

(v_{yf} - v_{yi})/a = t

Plugging in the numbers we get:

t = (0 - 35)/(-32.17)

t = 1.088 seconds

However, the ball still needs to fell back to Earth and will take the exact same time to fall down as it did to rise up. So the ** total time** is 1.088(2) =

__2.176 seconds__Now, considering ONLY the x-direction, the velocity in the x-direction can be calculated using cos(30°) = adjacent/hypotenuse = v_{x}/70 therefore v_{x} = 70cos(30°) = 60.6218 ft/s. It travels at 60.6218 ft/s for 2.176 seconds (as calculated above) so the ** distance it travels** is (60.6218)(2.176) =

__131.9 ft__