A cannon launches the cannon ball to the maximum horizontal distance of 3279 ft. Assuming no air resistance, and assuming the cannon ball is launched at ground level, what is the minimum initial speed of the cannon ball (just as it leaves the cannon) that is needed for it to reach this distance, in m/s

The minimum initial speed of the cannon ball is approximately 99.018 m/s

Explanation:

The parameters of the cannon are;

The maximum horizontal distance reached, R = 3,279 ft. ≈ 999.4392 m

The maximum horizontal range, , is given by the following formula;

Where;

u = The initial speed

g = The acceleration due to gravity ≈ 9.81 m/s²

= 999.4392 m

We get;

u = √(g × )

∴ u = √(9.81 m/s² × 999.4392 m) ≈ 99.018 m/s

The minimum initial speed of the cannon ball (just as it leaves the cannon) that is needed for it to reach this distance, in m/s, u ≈ 99.018 m/s.

c. magnitude and direction

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west