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Special Cases of Force Projectile Motion

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Projectile Motion A projectile is any object which once a force is used to throw it, hit it , propel it in some fashion, no other force acts on the object except for gravity. Projection can be horizontal, with no initial vertical velocity vertical, with some initial vertical velocity

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**Projectile Motion: A special case of uniformly accelerated motion**

This path is a parabola. Projectile Motion: A special case of uniformly accelerated motion If air resistance is negligible then only gravity affects the path (trajectory) of a projectile.

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**Characteristics of Projectile Motion**

The motion’s dimensions are vectors. The projectile will move along it’s trajectory (path) in both the horizontal (x) direction and the vertical (y) direction at the same time. The two directions are independent of each other. Only time is the same between the horizontal and vertical motion. It is, after all, only one object.

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Independent Motions Horizontal Vertical

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**Real Motion is the Combination of the Horizontal and Vertical Motions**

The blue dot show the real motion The path taken is the trajectory This is horizontal projection

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**Initial Velocity – V Along the Trajectory**

Angular projection - The initial velocity is the resultant of adding the two vector quantities together. The projection includes an angle of projection

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**Initial Velocity has an X and Y Component**

The vertical component and the horizontal component are independent of each other.

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**Horizontal and Vertical Components of Velocity**

Vertical velocity decreases at a constant rate due to the influence of gravity. It becomes zero. Then increases in the negative direction Negative velocity gets larger Vertical velocity = 0 Positive velocity gets smaller

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**Calculating Components**

You have learned to calculate components of a vector when we looked at inclined planes. The components are calculated by using the trig functions. The initial velocity acts as the hypotenuse of the right triangle. Vi

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**Continued Calculation**

The vertical velocity and the horizontal velocity are the legs of the triangle. Vy Vx Vi Calculate the Vy using sin = o/h Calculate Vx using cos = a/h

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**Horizontal Displacement of Projectiles**

Horizontal projection or projection at an angle graph of horizontal displacement v time As time progresses, the object gets further and further from its starting point. Time Displacement

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**Vertical Displacement of Projectiles**

Vertical Projection and Projection at an Angle graph of vertical displacement (height) v time The object rises into the air and then return to earth. Time Height

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**Vertical Displacement of Projectiles**

Horizontal projection Graph of vertical displacement (height) vs time The object leaves a horizontal surface and fall to the ground. Time Height

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**Horizontal Velocity of Projectiles**

Horizontal Projection and Projection at an Angle Time Velocity The horizontal velocity of a projectile remains constant from the time it is projected until gravity brings it to the ground. Remember: We are using an ideal situation where there is no air resistance

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**Vertical Velocity of Projectiles**

Vertical Projection and Projection at an Angle Time Velocity For objects projected directly upward or projected at some angle above the ground, the vertical velocity must begin positive, decrease to zero, and then increase in the negative direction. (Remember, gravity is negative)

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**Vertical Velocity of Projectiles**

Horizontal projection Time Velocity Horizontal projection begins with an initial vertical velocity of zero. The vertical velocity then increases in a negative direction.

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**Horizontal Acceleration of Projectiles**

Since we are idealizing the projection, we do not take into account any air resistance. We can, therefore, say there is no horizontal acceleration. Time Acceleration

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**Vertical Acceleration of Projectiles**

Vertical acceleration is the result of the pull of gravity, (-9.8 m/s2) This is the same on the way up and on the way down. Acceleration -9.8 Time

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**Important Notes on Vertical and Angular Projection**

The range (x) is the farthest the object will travel horizontally. The maximum height (ymax ) is the farthest the object will travel vertically. Y equals zero when it is at its lowest point.

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Determining the Range You can determine the displacement (range) of a projectile any any point along the trajectory. X = horizontal distance (range) vx = horizontal velocity t = time

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**Determining the Height**

You can determine the height (y) at any point in the trajectory! y = vertical displacement vy = initial vertical velocity g = acceleration due to gravity t = time

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**Other Important Notes on Vertical and Angular Projection**

At the highest point of the trajectory, it is the exact midpoint of the time. It takes the projectile half of the time to get to the top. When the projectile gets to the top, it has to stop going up and start going down, so the velocity in the y-direction at the highest point is zero for a split second. As the projectile falls, it is in free fall.

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**3 Primary Factors Affecting Trajectory**

Projection angle aka release angle or take-off angle Projection velocity aka initial or take-off velocity Projection height aka above or below landing

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Projection Angle The optimal angle of projection is dependent on the goal of the activity. For maximum height, the optimal angle is 90o. For maximum distance, the optimal angle is 45o.

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**The effect of Projection angle on the Range of a projectile**

10 degrees 30 degrees 40 degrees 45 degrees 60 degrees 75 degrees The angle that maximizes Range is = 45 degrees

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**The effect of Projection velocity on the Range of a projectile**

10 45 degrees Range ~ 10 m 20 45 degrees Range ~ 40 m 30 45 degrees Range ~ 90 m 100 90 80 70 60 50 40 30 20 10

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Projectile Problems Ignore air resistance. ay = g = m/s2 Set up the two dimension separately Origin x Origin y Positive x Positive y xi = constant yi = vxi = vyi = ax = ay = g

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**Projectile Problems – Two Dimensional Kinematics**

Write general kinematic equations for each direction Rewrite them for the problem at hand Find the condition that couples the horizontal and vertical motions (usually time)

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**Equations of Constant Acceleration**

This is the only equation to use for the horizontal part of the motion x = vxt d = ½ (vyf + vyi)t d = vyit + ½ gt2 vvf2 = vvi2 + 2gd These 3 equations are for the vertical part of the motion

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**The Monkey and the Banana**

A zookeeper must throw a banana to a monkey hanging from the limb of a tree. The monkey has a habit of dropping from the tree the moment that the banana is thrown. If the monkey lets go of the tree the moment that the banana is thrown,will the banana hit the monkey?

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**When you take gravity into consideration you STILL aim at the monkey!**

Monkey’s Gravity free path is “floating” at height of limb Banana’s Gravity free path Fall thru same height It works! Since both banana and monkey experience the same acceleration each will fall equal amounts below their gravity-free path. Thus, the banana hits the monkey.

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**Homework Chapter 6 read to page 152**

Problems: page 164, # 33,35,51,52,53,56 57,60.

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