Given the velocity of a projectile the angle of projection and the mass of the object how do you determine the distance traveled?

Answer 1

Given the initial velocity V, and the angle from the ground A, the total distance travelled X will be: X = 2 V2 cos(A) sin(A) / g

where "g" is the acceleration due to gravity, on earth g is approximately 9.81 m/s2.

You will notice that the mass of the object does not affect the distance traveled. We can derive this by first determining how long the projectile will be in the air. If the initial velocity is V, then the initial vertical velocity is Vsin(A). The vertical velocity will decrease at a rate of 'g' until the vertical velocity reaches zero (known as apogee), and the projectile starts falling down. The time from launch to apogee will be Vsin(A)/g.

The time for the projectile to go up is the same as for the projectile to fall down again, so the total time in the airis 2Vsin(A)/g.

Assuming we neglect friction, the horizontal velocity is Vcos(A) and does not change. The total distance traveled horizontally is the horizontal speed multiplied by the time spend in the air. So X = 2Vsin(A)/g * Vcos(A) = 2V2cos(A)sin(A)/g.

The maximum distance is achived with an angle of 45o. The distance travelled is symmetric around this value, i.e. an angle of 50o will give the same distance as 40o, and an angle of 15owill give the same distance as 75o.