projectiles solutions

rogers (nr9358) Projectiles volle (301) 1

This print-out should have 24 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.

001 10.0 pointsAt what point in its trajectory does a battedbaseball have its minimum speed?

1. at the top correct

2. at the beginning point

3. at the end point

4. somewhere at the middle height

Explanation:Since the horizontal component of velocity

of the batted baseball remains unchanged inits trajectory, the baseball has its minimumspeed when its vertical component of velocityis zero; i.e. at the top of its trajectory.

002 10.0 pointsA heavy crate accidentally falls from a high-flying airplane just as it flies directly above ashiny red Camaro parked in a parking lot.Relative to the Camaro, where will the

crate crash?

1. The crate will hit the Camaro.

2. The crate will hit the front part of thecar.

3. The crate will not hit the Camaro, butwill crash a distance beyond it determined bythe height and speed of the plane. correct

4. The crate will continue to fly and will notcrash.

Explanation:The crate starts falling when it is directly

above the camaro. It has a horizontal velocityequal to that of the airplane. Under theinfluence of gravity it will start acceleratingdown while keeping its horizontal velocity.It will finally crash a distance beyond the

camaro which is determined by the heightand speed of the plane.

003 10.0 points

A ball is thrown and follows the parabolicpath shown. Air friction is negligible. PointQ is the highest point on the path. Points Pand R are the same height above the ground.

Q

RP

How do the speeds of the ball at the threepoints compare?

1. ~vQ < ~v

P = ~v

R correct

2. ~vR < ~v

Q < ~v

P

3. ~vP < ~v

Q < ~v

R

4. ~vP = ~v

R = ~v

Q

5. ~vQ < ~v

R < ~v

P

6. ~vP = ~v

R < ~v

Q

Explanation:The speed of the ball in the x-direction is

constant. Because of gravitational accelera-tion, the speed in the y-direction is zero atpoint Q. Since points P and R are located atthe same point above ground, by symmetrywe see that they have the same vertical speedcomponent (though they do not have the samevelocity). The answer is then v

Q< v

P= v

R.

004 10.0 points

A ball is thrown and follows the parabolic


path shown. Air friction is negligible. PointQ is the highest point on the path. Points Pand R are the same height above the ground.

Q

RP

Which of the following diagrams best indi-cates the direction of the acceleration, if any,on the ball at point P?

1. The ball is in free-fall and there is noacceleration at any point on its path.

2.

correct

3.

4.

5.

6.

7.

8.

9.

Explanation:Since air friction is negligible, the only ac-

celeration on the ball after being thrown is

that due to gravity, which acts straight down.

005 (part 1 of 2) 10.0 pointsJanet jumps off a high diving platform witha horizontal velocity of 2.66 m/s and lands inthe water 2.3 s later.How high is the platform? The acceleration

of gravity is 9.8 m/s2 .

Correct answer: 25.921 m.

Explanation:

Let : v = 2.66 m/s ,

t = 2.3 s , and

g = 9.8 m/s2 .

Since vo = 0, the height of the fall is

h = vo t+1

2g t2 =

1

2g t2

=1

2(9.8 m/s2)(2.3 s)2 = 25.921 m .

006 (part 2 of 2) 10.0 pointsHow far from the base of the platform doesshe land?


Explanation:The horizontal velocity is constant, so the

horizontal distance is

d = v t = (2.66 m/s)(2.3 s) = 6.118 m .

007 10.0 pointsAssume: A 78 g basketball is launched at anangle of 58.3 and a distance of 19.8 m fromthe basketball goal. The ball is released at thesame height (ten feet) as the basketball goalsheight.A basketball player tries to make a long

jump-shot as described above.The acceleration of gravity is 9.8 m/s2 .What speed must the player give the ball?

Correct answer: 14.7312 m/s.

Explanation:


Basic concepts: Horizontally,

voh = v cos

vh = voh

d = voh t

Vertically,vov = v sin

vv = vov g t

h = vov t 12g t2 .

Solution: At the maximum range of the ball,vfv = vov, so,

vov = vov g t

2 vov = g tt = 2

vovg

.

The maximum distance covered is

d = voh t =2 voh vov

g

d =2 v cos v sin

g

d =v2 (2 sin cos )

g=

v2 sin(2 )

g.

Thus the initial velocity is

v =

d g

sin[2 ]

=

(19.8 m) (9.8 m/s2)

sin[2 (58.3)]

= 14.7312 m/s .

008 10.0 pointsA 0.94 kg rock is projected from the edge ofthe top of a building with an initial velocity of11.3 m/s at an angle 60 above the horizontal.Due to gravity, the rock strikes the ground ata horizontal distance of 15.8 m from the baseof the building.

Building

15.8 m

6011

.3m/s

h

How tall is the building? Assume theground is level and that the side of the build-ing is vertical. The acceleration of gravity is9.8 m/s2 .


Explanation:

Let : = 60 ,

v0 = 11.3 m/s ,

x = 15.8 m , and

m = 0.94 kg .

The flying time can be determined by

x = v0x t

t =x

v0x

=x

v0 cos .

From the point where the rock was projected(set to be the origin O), the y-coordinate ofthe point where the rock struck the ground is

y = v0y t 12g t2

= v0 sin

(x

v0 cos

) 1

2g

(x

v0 cos

)2

= x tan g x2

2 (v0 cos )2, so

h = |y|


=g x2

2 (v0 cos )2 x tan

=(9.8 m/s2) (15.8 m)2

2 [(11.3 m/s) cos 60]2

(15.8 m) tan 60= 10.9525 m .

009 10.0 pointsDuring a baseball game, a batter hits a pop-up to a fielder 77 m away.The acceleration of gravity is 9.8 m/s2 .If the ball remains in the air for 6.5 s, how

high does it rise?


Explanation:The distance to the fielder is extraneous

information and can be ignored.Consider the motion as the ball rises to its

maximum height h t =t

2. The height is

defined by

h =1

2g (t)2 =

1

2gt2

4=

g t2

8.

010 10.0 pointsA brick is thrown upward from the top of abuilding at an angle of 13.6 above the hori-zontal and with an initial speed of 17.6 m/s.The acceleration of gravity is 9.8 m/s2 .If the brick is in flight for 3.2 s, how tall is

the building?


Explanation:Basic ConceptThe height of the building is determined by

the vertical motion with gravity acting downand an initial velocity acting upward:

y = y0 + v0y t 12g t2

SolutionChoose the origin at the base of the build-

ing. The initial position of the brick is y0 = h,the vertical component of the initial velocity

is v0y = v0 sin directed upward, and y = 0when the brick reaches the ground, so

0 = h+ v0y t 12g t2

h = v0y t+ 12g t2

= (4.1385 m/s) (3.2 s)+

1

2

(9.8 m/s2

)(3.2 s)2

= 36.9328 m .

011 (part 1 of 3) 10.0 pointsA projectile of mass 0.352 kg is shot from acannon. The end of the cannons barrel isat height 6.7 m, as shown in the figure. Theinitial velocity of the projectile is 11 m/s .The projectile rises to a maximum height of

y above the end of the cannons barrel andstrikes the ground a horizontal distance xpast the end of the cannons barrel.

x

11m/s

51

y

6.7m

Find the time it takes for the projectile toreach its maximum height. The accelerationof gravity is 9.8 m/s2 .

Correct answer: 0.872307 s.

Explanation:

Let : vi = 11 m/s ,

= 51 , and

g = 9.8 m/s2 .

Vertically,

vyi = vi sin

= (11 m/s) sin 51

= 8.54861 m/s and


vyf = vyi g t .Since vytop = 0 ,

0 = vyi g t1t1 =

vyig

=8.54861 m/s

9.8 m/s2

= 0.872307 s .

012 (part 2 of 3) 10.0 pointsHow long does it take the projectile to hit theground?


Explanation:

Let : (xi, yi) = (0 m, 6.7 m) .

For the vertical motion,

v2yf = v2

yi 2 gy .Since vytop = 0 ,

0 = v2vi 2 gy

y =v2yi2 g

=(8.54861 m/s)2

2 (9.8 m/s2)

= 3.7285 m and

the time to fall from the top to the ground is

t2 =

2 (y + yi)

g

=

2 (3.7285 m + 6.7 m)

9.8 m/s2

= 1.45886 s ,

so the total time is

t = t1 + t2

= 0.872307 s + 1.45886 s

= 2.33116 s .

013 (part 3 of 3) 10.0 pointsFind the range x of the projectile.


Explanation:The horizontal component of vector vi is

vi =vxicos

vxi = vi cos

= (11 m/s) cos 51

= 6.92252 m/s .

The horizontal motion has no acceleration,so

x = vxi t

= (6.92252 m/s) (2.33116 s)

= 16.1375 m .

Alternate Solution: y = 0 when theprojectile hits the ground, so

y = yi + vyi t1

2g t2

= yi +vyivxi

x g2

x2

v2xi= 0

(g

2 v2xi

)x2

(vyivxi

)x+ yi = 0

Since

a =g

2 v2xi=

9.8 m/s2

2 (6.92252 m/s)2

= 0.102251 m1 ,

b = vyivxi

= 8.54861 m/s6.92252 m/s

= 1.2349 ,

c = yi = 6.7 m , and

b2 4 a c = (1.2349)2 4 (0.102251 m1) (6.7 m)

= 4.26529 ,

x =b b2 4 a c

2 a

=1.23494.265292 (0.102251 m1)

= 16.1375 m .


014 (part 1 of 3) 10.0 pointsAn Alaskan rescue plane traveling 38 m/sdrops a package of emergency rations froma height of 199 m to a stranded party ofexplorers.The acceleration of gravity is 9.8 m/s2 .Where does the package strike the ground

relative to the point directly below where itwas released?


Explanation:

Basic ConceptsProjectile motion with an initial horizontalvelocity.

SolutionHorizontally, there is no acceleration, so

x = xo + voxt

vf = vo = v

Vertically,

y = yo + voyt 12gt2

vy = voy gtConsider the point of release to be the originof the coordinate system.The time for the package to reach the groundis determined by the vertical motion wherevoy = 0, y = h, and a = g:

2(h) = gt2

t =

2h

g

Horizontally, xo = 0 and a = 0, so

x = vt

015 (part 2 of 3) 10.0 pointsWhat is the horizontal component of the ve-locity just before it hits?

Correct answer: 38 m/s.

Explanation:There is no horizontal acceleration, so the

horizontal velocity is constant.

016 (part 3 of 3) 10.0 pointsWhat is the vertical component of the velocityjust before it hits? (Choose upward as thepositive vertical direction)

Correct answer: 62.4532 m/s.Explanation:vo = 0, so

vy = gt

017 10.0 pointsA car is parked near a cliff overlooking theocean on an incline that makes an angle of34.5 with the horizontal. The negligent driverleaves the car in neutral, and the emergencybrakes are defective. The car rolls from restdown the incline and has a velocity 5 m/swhen it reaches the edge of the cliff. The cliffis 39.7 m above the ocean.

How far is the car from the base of the cliffwhen the car hits the ocean? The accelerationof gravity is 9.8 m/s2 .


Explanation:First, find the cars initial vertical velocity

when it leaves the cliff

v0y = v sin

= (5 m/s) sin(34.5)

= 2.83203 m/s .


Then find the vertical velocity with which thecar strikes the water as

v2y = v2

0y + 2 g h

= (2.83203 m/s)2 + 2 (9.8 m/s2) (39.7 m)

= 786.14 m2/s2

vy = 28.0382 m/s .

For the time of flight,

vy = v0y + g t

t =vy v0y

g

=28.0382 m/s 2.83203 m/s

9.8 m/s2

= 2.57206 s .

The initial velocity is v0x = v cos and thehorizontal motion of the car during this timeis

x = v0x t = (5 m/s) cos(34.5) (2.57206 s)

= 10.5985 m .

018 10.0 pointsA plane drops a hamper of medical suppliesfrom a height of 3460 m during a practice runover the ocean. The planes horizontal veloc-ity was 149 m/s at the instant the hamperwas dropped.What is the magnitude of the overall ve-

locity of the hamper at the instant it strikesthe surface of the ocean? The acceleration ofgravity is 9.8 m/s2 .

Correct answer: 300.028 m/s.

Explanation:This is a projectile motion problem. The

motion of the dropping hamper consists of twoparts: horizontally, it moves with the initialvelocity of the plane, i.e., vh = v = 149 m/s ;vertically, due to gravity, it moves as a freelyfalling body. Thus

v2 = 2 g h

vv =2 g h =

2 (9.8 m/s2) (3460 m)

= 260.415 m/s

and the overall velocity at the instant thehamper strikes the surface of the ocean is

vf =v2v + v

2

h

=(260.415 m/s)2 + (149 m/s)2

= 300.028 m/s .

019 10.0 pointsSomeone in a car going past you at the

speed of 28 m/s drops a small rock from aheight of 1.5 m.How far from the point of the drop will the

rock hit the ground? The acceleration due togravity is 9.8 m/s2.


Explanation:

Let : v = 28 m/s ,

h = 1.5 m , and

g = 9.8 m/s2 .

Vertically, the rock accelerates downwardlike any other falling object. The verticalmotion defines the time of the fall:

h = v0t+

1

2g t2 =

1

2g t2

since vo = 0, so

t =

2h

g

=

2 (1.5 m)

9.8 m/s2

= 0.553283 s.

Horizontally, the rock moves at constantvelocity (with no acceleration), so

d = v t

= (28 m/s) (0.553283 s)

= 15.4919 m .


.

020 (part 1 of 2) 10.0 pointsAn artillery shell is fired at an angle of 30.8

above the horizontal ground with an initialspeed of 1660 m/s.The acceleration of gravity is 9.8 m/s2 .Find the total time of flight of the shell,

neglecting air resistance.

Correct answer: 2.89113 min.

Explanation:Using half the flight we have

v0y = v0 sin 0

and0 = vfy = v0 sin 0 + a0yt,

so that

t =v0 sin 0

g.

The total time of flight, by symmetry, is

ttot =2v0 sin 0

g

=2(1660 m/s) (sin 30.8)

9.8 m/s2

= 2.89113 min .

021 (part 2 of 2) 10.0 pointsFind its horizontal range, neglecting air resis-tance.

Correct answer: 247.343 km.

Explanation:This part can be worked out without using

the numerical value of the time of flight. Theinitial components of the initial velocities are

v0x = v0 cos 0

andv0y = v0 sin 0,

so thatx = v0 cos 0 t

t =x

v0 cos 0

and

y = v0 (sin 0) t 12g t2.

Substituting t into the last equation yields

y =sin 0cos 0

x g2 v2

0(cos2 0)

x2.

At the target y = 0 and x = R (the range), so

0 =sin 0cos 0

gR2 v2

0(cos2 0)

or

R =2 v2

0(sin 0) (cos 0)

g

=2 (1660 m/s)2 (sin 30.8) (cos 30.8)

9.8 m/s2

= 2.47343 105 m = 247.343 km .

022 (part 1 of 3) 10.0 pointsA ball of mass 0.3 kg, initially at rest, iskicked directly toward a fence from a point20 m away, as shown below.The velocity of the ball as it leaves the

kickers foot is 17 m/s at angle of 47 abovethe horizontal. The top of the fence is 6 mhigh. The ball hits nothing while in flight andair resistance is negligible.The acceleration due to gravity is 9.8 m/s2.

b

20 m

6 m

17m/s

47

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

b

Determine the time it takes for the ball toreach the plane of the fence.


Explanation:

Let : = 47 and

d = 20 m .


The horizontal component of the velocity isconstant, so

vhoriz = v0 cos

= (17 m/s) cos 47

= 11.594 m/s .

The horizontal motion defines the time offlight:

vhoriz t = d

t =d

vhoriz

=20 m

11.594 m/s

= 1.72503 s .

023 (part 2 of 3) 10.0 pointsHow far above the top of fence will the ball

pass? Consider the diameter of the ball to benegligible.


Explanation:The vertical component of the initial veloc-

ity is

vvert = v0 sin

= (17 m/s) sin 47

= 12.433 m/s .

The height of the ball during its flight isgiven by

y = vvert t 12g t2

= (12.433 m/s) (1.72503 s)

12(9.8 m/s2) (1.72503 s)2

= 6.86623 m .

Therefore, the distance that the ball passesabove the fence is

y = (6.86623 m) (6 m) = 0.866231 m .

024 (part 3 of 3) 10.0 points

What is the vertical component of the velocitywhen the ball reaches the plane of the fence?

Correct answer: 4.47232 m/s.Explanation:Thus the the vertical component of the ve-

locity when the ball reaches the plane of thefence is

vvert = v0 sin g t= (17 m/s) sin 47

(9.8 m/s2) (1.72503 s)= 4.47232 m/s .

This is verified by analyzing the graph below

0 0.5 1.0 1.5 2.0 2.52520151050

5

10

15

20

25

Time (s)

VerticalComponentofVelocity

(m/s)

Vertical Velocity vs Time

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