Tutorial for Projectile Motion Tutorial for Projectile Motion ProblemsProblems

Components of VelocityComponents of VelocityIf an object is launched at an angle, If an object is launched at an angle, its velocity is not perfectly horizontal its velocity is not perfectly horizontal (x direction), nor is it perfectly (x direction), nor is it perfectly vertical (y direction).vertical (y direction).An object launched at an angle will An object launched at an angle will have a velocity that is a combination have a velocity that is a combination of the Vx and the Vy.of the Vx and the Vy.

You must always find the You must always find the components of the initial velocity.components of the initial velocity.If the velocity is given, you must use If the velocity is given, you must use the angle to determine what part of the angle to determine what part of the initial velocity is in the X the initial velocity is in the X direction and what part is in the Y direction and what part is in the Y direction.direction.

Find V1y and VxFind V1y and VxV0y=V0sin V0y=V0sin ΘΘ (this is the equation (this is the equation for V0y always)for V0y always)Vx=V0 cos Vx=V0 cos ΘΘThese are known as the These are known as the “components” of the velocity.“components” of the velocity.

Changes in VelocityChanges in VelocityThe velocity in the Y direction must The velocity in the Y direction must change. The velocity in the Y change. The velocity in the Y direction decreases because any direction decreases because any object that goes up must be affected object that goes up must be affected by gravity.by gravity.The velocity in the X direction does The velocity in the X direction does not change at all, it is the same not change at all, it is the same EVERYWHERE along the path.EVERYWHERE along the path.

Looking for half timeLooking for half timeThe time it takes for an object to reach The time it takes for an object to reach its maximum height will be half of the its maximum height will be half of the time it spends in the air.time it spends in the air.At the maximum height, the velocity in At the maximum height, the velocity in the Y direction will be 0 m/sthe Y direction will be 0 m/s

2 1y yhalf

v vt

g

2 1y yhalf

v vt

g

2 1y yhalf

v vt

g

2 1y yhalf

v vt

g

2 1y yhalf

v vt

g

1 0y yhalf

v vt

g

Total TimeTotal TimeThe total time an object is in the air The total time an object is in the air is simply twice the half time.is simply twice the half time.The total time is used to find the The total time is used to find the maximum range (X) of the object maximum range (X) of the object (measured in meters).(measured in meters).

Finding Max HeightFinding Max HeightTo find maximum height, use the following To find maximum height, use the following equation:equation:The time in this equation must be the half The time in this equation must be the half time!time!

20

12yy v t gt

To Find RangeTo Find RangeTo find the maximum range, you must use To find the maximum range, you must use the total time in the following equation:the total time in the following equation:

xx v t

Notes about solving problemsNotes about solving problemsIf the equation calls for V0y, you If the equation calls for V0y, you must use the V0y value you must use the V0y value you calculated using V0sincalculated using V0sinΘΘ..If the equation calls for Vx, you must If the equation calls for Vx, you must use the Vx value you calculated use the Vx value you calculated using V0cos using V0cos ΘΘ..Please note that Vx, V0y, and V0 are Please note that Vx, V0y, and V0 are not the same value!!! not the same value!!!

Finding the V1 at a specific timeFinding the V1 at a specific timeIf you are asked to find the V1 at a If you are asked to find the V1 at a specific time, you must first find the specific time, you must first find the components of the initial velocity (Vx and components of the initial velocity (Vx and V0y).V0y).Next, find the V1y at the given time using Next, find the V1y at the given time using the below equation:the below equation:

1 0y yv v gt

Finding V1 continuedFinding V1 continuedOnce you have V1y at the given time, Once you have V1y at the given time, and you have Vx, simply use the and you have Vx, simply use the Pythagorean Theorem to find V1:Pythagorean Theorem to find V1:

2 21 1x yv v v

Finding the Angle at Any pointFinding the Angle at Any pointTo find the angle at any point, first find the To find the angle at any point, first find the Vx and the Vy at that point.Vx and the Vy at that point.Once those are determined, use the Once those are determined, use the tangent trig function.tangent trig function.ΘΘ=Tan=Tan-1-1 (Vy/Vx) (Vy/Vx)

Example ProblemsExample Problems1.1. A ball is kicked into the air at 30 m/s at an angle A ball is kicked into the air at 30 m/s at an angle

of 25 degrees. What is its maximum range and of 25 degrees. What is its maximum range and height?height?

1 0y yhalf

v vt

g

Given:V1: 30 m/sΘ=25 degreesV2y=0 m/sg= -9.8 m/s2

Find:VxV1ythttXY

Vx=V0 cos Θ=30 cos 25= 27.19 m/sV0y=V0 sin Θ= 30 sin 25=12.68 m/s

21(12.68)(1.29) ( 9.8)(1.29 ) 8.202

y m

Ttotal=2.59 s

0 12.68 1.299.8halft s

xx v t (27.19)(2.59) 70.36x m

20

12yy v t gt

Example ProblemsExample ProblemsA rocket is launched from the ground and is aimed 2500 m A rocket is launched from the ground and is aimed 2500 m away. If it reaches a maximum height of 200 meters, how away. If it reaches a maximum height of 200 meters, how fast must it be going when it leaves the ground (find Vy and fast must it be going when it leaves the ground (find Vy and Vx)? At what angle must it be launched? (Hint: find time Vx)? At what angle must it be launched? (Hint: find time first using cliff problems)first using cliff problems)

2 20 0x yv v v

Given:X: 2500 mY: 200 m

Find:V1yVxthttV1

To find the initial velocity (V0), find Vx and V0y. First find time using the initial velocity in the y direction as zero.

1/ 2ytg

= 6.39sT Total=12.78s

V0y= V1y-gt = 62.62 m/s

Vx= x = 195.62 m/s t

2 20 195.62 62.62 205.40 /v m s

3. A basketball player jumps up at 15 3. A basketball player jumps up at 15 m/s at an angle of 50 degrees. At m/s at an angle of 50 degrees. At what time will he be 2 meters away what time will he be 2 meters away (horizontally) from where he (horizontally) from where he jumped?jumped?

Given:V0: 15 m/sΘ=50 degreesX=2 m/s

Find:t

Vx=V0 cos Θ=9.64 m/s

T=x= .21 s Vx

4. A football is kicked with an initial 4. A football is kicked with an initial velocity of 55 m/s at an angle of 60 velocity of 55 m/s at an angle of 60 degrees. At what times will it pass degrees. At what times will it pass 5 meters above the ground? (you 5 meters above the ground? (you can use quadratic formula in the can use quadratic formula in the calculator)calculator)

Θ

Given:V0: 55 m/sΘ=60 degreesY=5 m

Find:T1T2

V0y= V0 sin Θ= 47.63 m/s

The ball will pass 2 meters high twice.You can use the quadratic formula to find the two times

ax2+ bx + c=0

Make equation look like this1/2gt2+V1yt-y=0

T1=.11sT2= 9.61s

5. A baseball is thrown into the air with a velocity of 5. A baseball is thrown into the air with a velocity of 27.5 m/s at an angle of 45 degrees. What will its 27.5 m/s at an angle of 45 degrees. What will its velocity be after 3 seconds? (Hint: find its Vy and velocity be after 3 seconds? (Hint: find its Vy and Vx at 3 seconds, use the Pythagorean Theorem).Vx at 3 seconds, use the Pythagorean Theorem).

2 21 1x yv v v

Given:V0: 27.5 m/sΘ=45 degreesT= 15 sec.

Find:V1V1yVxV0y

Vy= V0 sin Θ=19.45 m/sVx= V0 cos Θ= 19.45 m/s

V1y=V0y+gt=9.65 m/s

2 21 19.45 .9.65 21.71 /v m s

6. A rocket is shot up into the air. If it 6. A rocket is shot up into the air. If it has a range of 1000 m, and reaches has a range of 1000 m, and reaches a height of 800 m, what angle was a height of 800 m, what angle was it shot up at and what was its initial it shot up at and what was its initial velocity? velocity?

Givens:X=1000 mY= 800 mG=9.8 m/s2

Find:V0V0yVxthtt

You must first find the half way time, how long it the rocket to drop the 800 m; set V1y equal to zero for this.

1/ 2ytg

800 12.78

1/ 2(9.8)t s

Next, find the total time by doubling the half way timeTtotal= 25.56 seconds

Find Vx using the given distance (x) and the total time

xxvt

1000 39.13 /25.56xv m s

#6 Continued#6 ContinuedFind V1y by setting V2y equal to zero and using the half way Find V1y by setting V2y equal to zero and using the half way

timetime

0 1 0 ( 9.8)(12.78) 125.24 /y yv v gt m s Find the V1 by using the Pythagorean Theorem with V1y and Vx.

Find the angle that it is launched at by using the tangent function

1 125.24tan ( ) 72.65deg39.13

2 20 0x yv v v 2 2

0 39.13 125.24 131.21 /v m s

01tan ( )yVVx

7. At what times will a rocket launched at an angle of 7. At what times will a rocket launched at an angle of 37 degrees and a velocity of 400 m/s pass 100 m 37 degrees and a velocity of 400 m/s pass 100 m high? Use Quadratic formula for this.high? Use Quadratic formula for this.

0 0 sinyv v Given:Θ=37 degreesV0= 400 m/sY=100 m

Find:t1t2V1y

Find V1y first:

To find the two times that the rocket passes that height, you must use the Quadratic formula: Set y=V0yt+1/2 gt2 to look like ax2+ bx+c=0. You are solving for t, so your A value is -4.9, your B value is 240.73 m/s (V0y), and your C value is -100 . Use the calculator to plug in and find the two times it passes those two heights.

T1= .42 st2= 48.71 s

0 400(sin37) 240.73 /yv m s