*This specific example can be easily adapted for any initial velocity and any angle of elevation.*

A projectile is an object that is given an initial velocity, and is acted on by gravity.

The path the object follows is determined by these effects (ignoring air resistance). The trajectory has horizontal (x) and vertical (y) components.

The projectile’s trip can be broken into two parts: going up and coming The range is the point where the projectile strikes the ground which happens at the time we found in Part b of the problem.

R = 106.1 m/s · 21.6s R = 2291.8 m The projectile landed 2291.8 meters from the canon. The position has two components: horizontal and vertical position.

The horizontal and vertical velocities are expressed in meters per second (m/s).

Horizontal distance horizontal distance = (initial horizontal velocity)(time) x = v The velocity of the ball after 5.00 s has two components.Once these two components are found, they must be combined using vector addition to find the final velocity.The ball was kicked horizontally, so v = -49.0 m/s In projectile motion problems, up is defined as the positive direction, so the y component has a magnitude of 49.0 m/s, in the down direction.To find the magnitude of the velocity, the x and y components must be added with vector addition: v ∴v = 51.24 m/s The magnitude of the velocity is 51.24 m/s.Though it was not asked for in the question, it is also possible to find the direction of the velocity as an angle.If you need to know the velocity of the projectile at a specific time, you can use the formulav – v = atand solve for v.Just remember velocity is a vector and will have both x and y components.Therefore, the amount of time that the toy rocket spent in the air was 3.53 seconds.An object is thrown straight up from the top of a building h feet tall with an initial velocity of v feet per second.F = ma = -mgsolve for aa = -g Now we have enough information to find the time.We know the initial vertical velocity (V h = 1145.9 m – 571.5 m h = 574.4 m The highest height the projectile reaches is 574.4 meters. We’ve already done most of the work to get this part of the question if you stop to think.

## Comments How To Solve A Projectile Motion Problem

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