# The projectile flew out of the cannon with an initial speed of 800m / s at an angle of 30

**The projectile flew out of the cannon with an initial speed of 800m / s at an angle of 30 to the horizon. Determine the highest altitude and range.**

V0 = 800 m / s.

∠α = 30 “.

g = 9.8 m / s ^ 2.

h -?

L -?

Horizontally, the projectile moves uniformly at a speed Vx = V0 * cosα. The flight range L is found by the formula: L = Vх * t = V0 * cosα * t, where t is the projectile flight time.

Vertically, the projectile moves uniformly with free fall acceleration g and initial velocity V0у = V0 * sinα.

h = (V0у) ^ 2/2 * g = (V0 * sinα) ^ 2/2 * g.

h = (800 m / s * sin30 “) ^ 2/2 * 9.8 m / s ^ 2 = 8163 m.

The flight time is expressed by the formula: t = 2 * V0у / g = 2 * V0 * sinα / g.

L = V0 * cosα * 2 * V0 * sinα / g = V0 ^ 2 * sin (2 * α) / g.

L = (800 m / s) ^ 2 * sin (2 * 30 “) / 9.8 m / s ^ 2 = 56,555 m.

Answer: h = 8163 m, L = 56555 m.