PHILIPPINE SCIENCE HIGH SCHOOL – Main CampusPhysics 2 – Advanced Topics in Physics I

LONG TEST 3: PROJECTILE MOTION

I. True or FalseDirection:

Write T if the statement is true and F if it is false. Note that air resistance is neglected in all statements.

01. A projectile fired at any angle above the ground has a constant horizontal velocity in all points.

02. The time in going up of a projectile is the same to its time in going down back to its initial level.

03. The range of a projectile is maximum if it is fired at angle of 45o.

04. A projectile fired at an angle of 25o covers the same range if it is fired at an angle of 65o with greater initial speed.

05. The vertical motion of a projectile is a uniformly accelerated motion.

06. A projectile is an object launched into space under the influence of gravity only.

07. A football is kicked at an angle of θ with respect to the horizontal. The acceleration is -9.8 m/s2 at all times.

08. At the highest point of the trajectory, the ball’s velocity is perpendicular to its acceleration.

09. A projectile fired horizontally will strike the ground in the same time as one dropped vertically from the same position if we neglect the effects of air resistance.

10. The range of a projectile is affected by its initial speed only.

II. PROBLEM SOLVINGDirection:

Solve all the problems completely and correctly in your answer sheet. Always follow the format in problem solving: Illustration (if necessary) → Given → Find → Derivation → Solution → Final answer → Box your final equation and answer. Follow significant figures rules.

Problems 01 to 04 pertain to the statement below:A projectile is fired at an angle of 60.0o above the

horizontal/ground with an initial speed of 30.0 m/s.01. (2 points) What is the initial horizontal velocity, vix of the

projectile?02. (2 points) What is the initial vertical velocity, viy of the

projectile?

03. (3 points) How far does the projectile travel horizontally (range), x before it hits the ground?

04. (3 points) How long (time) does it take the projectile to reach the highest point in its trajectory?

Problems 05 to 08 pertain to the statement below:A shell is fired horizontally in the positive x direction from

the top of an 80.0 m high cliff. The shell strikes the ground 1330 m from the base of the cliff.05. (3 points) Determine the initial velocity of the shell.06. (2 points) What is horizontal velocity of the shell as it hits

the ground? 07. (2 points) What is vertical velocity of the shell as it hits

the ground? 08. (3 points) What is the velocity (magnitude and direction)

of the shell as it hits the ground? Hint: use your answers in 06 and 07 and apply vector components

Problems 09 to 12 pertain to the statement below:A football is kicked with an initial speed vi at an angle θ

above the horizontal ground. It covers a horizontal distance/range of 26 m in 3.4 s.09. (2 points) What is horizontal velocity, vix of the football?10. (2 points) What is the initial vertical velocity, viy of the

football?11. (3 points) What is initial speed (magnitude of the

velocity) of the football? 12. (3 points) What is the angle of trajectory, θ? What is the

other angle that can give the same range of the football if it is kicked at the same speed?

GOD Bless!!! John 3:16Tract: dtquinsaat_07/31/2010/1832h

Test No.:_________

Answers and Solutions

I. True or False01. A projectile fired at any angle above the ground has a constant horizontal velocity in all points. – T 02. The time in going up of a projectile is the same to its time in going down back to its initial level. – T 03. The range of a projectile is maximum if it is fired at angle of 45o. – T 04. A projectile fired at an angle of 25o covers the same range if it is fired at an angle of 65o with greater initial speed. – F 05. The vertical motion of a projectile is a uniformly accelerated motion. – T 06. A projectile is an object launched into space under the influence of gravity only. – T 07. A football is kicked at an angle of θ with respect to the horizontal. The acceleration is -9.8 m/s2 at all times. – T 08. At the highest point of the trajectory, the ball’s velocity is perpendicular to its acceleration. – T 09. A projectile fired horizontally will strike the ground in the same time as one dropped vertically from the same position if we

neglect the effects of air resistance. – T 10. The range of a projectile is affected by its initial speed only. – F

II. PROBLEM SOLVINGProblems 01 to 04 pertain to the statement below:

A projectile is fired at an angle of 60.0o above the horizontal/ground with an initial speed of 30.0 m/s.01. (2 points) What is the initial horizontal velocity, vix of the projectile?02. (2 points) What is the initial vertical velocity, viy of the projectile?03. (3 points) How far does the projectile travel horizontally (range), x before it hits the ground?04. (3 points) How long (time) does it take the projectile to reach the highest point in its trajectory?

Illustration:

Given: v i=30.0m /s - initial velocity magnitude of the projectile

θ=60o - angle of projection

01. Find: v ix=? - initial horizontal velocity of the projectile

Derivation: v ix=v icosθSubstitution: v ix=(30.0m / s) cos60o v ix=15.0m /s

02. Find: v iy=? - initial vertical velocity of the projectile

Derivation: v iy=v isin θSubstitution: v ix=(30.0m / s) sin 60o v ix=26 .0m /s

03. Find: x=? - range

Derivation: x=v ix t=(v¿¿ icosθ) t ¿ where v y−v iy=at v y=0 at the highest point

So the total time is: t=−2v isin θ

a Therefore: x=

−2v i2cos θ sin θa

Substitution: x=−2 (30.0m /s )2 cos60osin 60o

−9.8m /s2x=79.5m

04. Find: t=? - time to reach the highest point of the trajectory

Derivation: v y−v iy=at v y=0 at the highest point

So the time is: t=−v i s∈θa

Substitution: t=−(30.0m /s )sin 60o

−9.8m/ s2t=2.65 s

Problems 05 to 08 pertain to the statement below:A shell is fired horizontally in the positive x direction from the top of an 80.0 m high cliff. The shell strikes the ground 1330 m

from the base of the cliff.

05. (3 points) Determine the initial velocity of the shell.06. (2 points) What is horizontal velocity of the shell as it hits the ground? 07. (2 points) What is vertical velocity of the shell as it hits the ground? 08. (3 points) What is the velocity (magnitude and direction) of the shell as it hits the ground? Hint: use your answers in 06 and 07

and apply vector componentsIllustration:Given: y=80.0m - height of the cliff

x=1330m - range

05. Find: v i=? - initial velocity of the shell

Derivation: v i=v ix=xt

where y=v iy t+12a t2 initially, v iy=0m / s

So, t=√ 2 ya Therefore, v i=

x

√ 2 ya v i=x√ a2 y

Substitution: v i=(1330m )√ −9.8m /s2

2 (−80.0m )v i=329m /s

06. Find: vx=? - horizontal velocity of the shell as it hits the ground

Derivation: horizontal velocity is constant. Therefore, vx=v i x=v iSubstitution: vx=329m /s

07. Find: v y=? - vertical velocity of the shell as it hits the ground

Derivation: v y2−v iy

2=2ay initially, v iy=0m / s so, v y=−√2aySubstitution: v y=−√2 (−9.8m /s2 ) (−80.0m ) v y=−39.6m /s

08. Find: v=? - velocity (magnitude and direction) as it hits the ground

Derivation: v=√v x2+v y2 so from the equations from 05 and 07: v=√ x2a2 y +4 ay

Simplifying, v=√a( x2+8 y22 y ) tanθ=v yvx

=−√2ay

x √ a2 y

=−√4 y2x

=−2 yx

θ=tan−1(−2 yx )Substitution: v=√ (−9.8m /s2 )[ (1330m )2+8 (−80.0m )2

2 (−80.0m ) ] v=334m / s

θ=tan−1(−2 (80.0m )1330m ) θ=−7o=7obelow horizontal

Problems 09 to 12 pertain to the statement below:A football is kicked with an initial speed vi at an angle θ above the horizontal ground. It covers a horizontal distance/range of 26

m in 3.4 s.09. (2 points) What is horizontal velocity, vix of the football?10. (2 points) What is the initial vertical velocity, viy of the football?11. (3 points) What is initial speed (magnitude of the velocity) of the football? 12. (3 points) What is the angle of trajectory, θ? What is the other angle that can give the same range of the football if it is kicked at

the same speed?

Illustration:

Given: x=26m - range reached by the football

t=3.4 s - time to cover the range or time of flight

09. Find: v ix=? - initial horizontal velocity of the football

Derivation: v ix=xt

Substitution: v i x=26m3.4 s

v i x=7.6m /s

10. Find: v iy=? - initial vertical velocity of the football

Derivation: y=v iy t+12a t2 back to initial level, y=0m so, 0=v iy t+

12a t 2

Therefore, v iy=−12at

Substitution: v iy=−12

(−9.8m /s2) (3.4 s ) v iy=17m / s

11. Find: v i=? - initial speed of the football

Derivation: v i=√v i x2+v i y2 so from the equations from 09 and 10: v i=√( xt )2

+(−12 at )2

Simplifying, v i=12 t

√4 x2+a2 t 4

Substitution: v i=[ 12 (3.4 s ) ]√4 (26m )2+(9.8m /s2)2 (3.4 s )4 v i=18m /s

13. Find: θ=? - angle of projection

Derivation: tanθ=v i yv i x

=

−12at

xt

=−a t 2

2 x therefore, θ=tan−1(−a t22x )

Substitution: θ=tan−1[−(−9.8m /s2 ) (3.4 s )2

2 (26m ) ] θ=65o=65oabovehorizontal