2.1.2 In Class or Homework Exercise
1. An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal
speed is 9.0 m/s as he leaves the ground, how long was he in the air and how
high did he go?
horizontal
d x  8.0m
vx  9.0m / s
t ?
d x
t
8.0
9.0 
t
t  0.89 s
vx 
To find out how high the jumper went, we must look at either the trip up or the
trip down. We will use the trip down so that the initial vertical velocity will be
zero.
vertical (using down as positive)
v yi  0
a y  9.80m / s 2
t  0.445s (half of the total time in the air)
d y  ?
d y  v yi t  12 a y t 2
 0  12 (9.80)(0.445) 2
 0.97m
2. An athlete throws the shot-put with an initial speed of 14 m/s at a 40.o angle to
the horizontal. The shot leaves the shot-putter’s hand at a height of 2.2 m
above the ground. Calculate the horizontal displacement travelled.
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v
vy
vx
v y  v sin 
vx  v cos 
 (14)(cos 40.)
 11m / s
 (14)(sin 40.)
 9.0m / s
horizontal
vertical (using up as positive)
vx  11m / s
v yi  9.0m / s
d x  ?
a y  9.80m / s 2
d y  2.2m
d y  v yi t  12 a y t 2
2.2  9.0t  12 (9.80)t 2
0  4.90t 2  9.0t  2.2
t1  0.22 s
t2  2.06 s
d x
t
d x
10.7 
2.06
d x  22m
vx 
3. A projectile is fired with an initial speed of 63.4 m/s at an angle of 24.6 o above
the horizontal on a long flat firing range. Determine
a. The maximum height reached by the projectile
v
vy
vx
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v y  v sin 
vx  v cos 
 (63.4)(cos 24.6)
 57.6m / s
 (63.4)(sin 24.6)
 26.4m / s
vertical (using up as positive)
v yi  26.4m / s
a y  9.80m / s 2
v yf  0
d y  ?
d y 

v yf2  v yi2
2a y
0  26.42
2(9.80)
 35.6m
b. The total time in the air.
v yf  v yi
ay 
t
0  26.4
9.80 
t
t  2.69 s
Since this is just the time to go up, the total time in the air is t  5.38s
c. The total horizontal distance covered.
horizontal
vx  57.5m / s
t  5.38s
d x  ?
d x
t
d x
57.5 
5.38
d x  309m
vx 
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d. The velocity of the projectile 1.50 s after firing.
The horizontal velocity of the projectile does not change. It will still be
vx  57.5m / s . We must find the new vertical velocity.
vertical (using up as positive)
v yi  26.4m / s
a y  9.80m / s 2
t  1.50s
v yf  ?
ay 
9.80 
v yf  v yi
t
v yf  26.4
1.50
v yf  11.7 m / s
vf
vyf
vxf
vf  v  v
2
xf
tan  
2
yf
v yf
vxf
11.7
57.5
  11.5
 (57.5) 2  (11.7) 2

 58.7m / s
v f  58.7m / s,11.5 above the horizontal
Notice that the projectile has slowed down and is more horizontal, as
one would expect before it reaches its highest point.
4. Trailing by two points, and with only 2.00 s remaining in a basketball game,
Pat makes a jump-shot at an angle of 60.0o with the horizontal, giving the ball
a velocity of 10.0 m/s. The ball is released at the height of the basket, 3.05 m
above the floor. YES! It's a score.
a. How much time is left in the game when the basket is made?
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v
vy
vx
v y  v sin 
vx  v cos 
 (10.0)(cos 60.0)
 5.00m / s
 (10.0)(sin 60.0)
 8.66m / s
vertical (using up as positive)
v yi  8.66m / s
a y  9.80m / s 2
d y  0 (since the ball is released at the same height as the basket)
t ?
d y  v yi t  12 a y t 2
0  8.66t  12 (9.80)t 2
tremaining  2.00  1.77
 0.23s
4.90t 2  8.66 t
t  1.77 s
b. The three-point line is a distance of 6.02 m from the basket. Did the Pat
tie the game or put his team ahead?
horizontal
vx  5.00m / s
t  1.77 s
d x  ?
d x
t
d x
5.00 
1.77
d x  8.85m
vx 
Since d x  6.02m , the basket is worth 3 points and he put his team
ahead.
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5. A cricket ball is hit at 30.0 m/s at an angle of 53.0o with the horizontal.
Immediately, a fielder runs 4.00 m/s toward the batter and catches the ball at
the same height it was hit. What was the original distance between the batter
and the fielder?
Ball
v
vy
vx
vx  v cos 
v y  v sin 
 (30.0)(cos 53.0)
 18.1m / s
 (30.0)(sin 53.0)
 24.0m / s
horizontal
vertical (using up as positive)
vx  18.1m / s
d y  0
d x  ?
v yi  24.0m / s
a y  9.80m / s 2
d y  v yi t  12 a y t 2
0  24.0t  12 (9.80)t 2
4.90t 2  24.0 t
t  4.90s
d xb
t
d xb
18.1 
4.90
d xb  88.7 m
vx 
Fielder
vx 
4.00 
d xf
t
d xf
4.90
d xf  19.6m
UNIT 2 2D Motion
dt  88.7  19.6
 108.3m
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6. A football is kicked at an angle of 37o with the horizontal with a velocity of
20.0 m/s. A fence is 31.0 m away and is 3.5 m high. Does the ball go over the
fence?
v
vy
vx
vx  v cos 
v y  v sin 
 (20.0)(cos 37)
 (20.0)(sin 37)
 16m / s
 12m / s
We will assume that the ball reaches the fence, and attempt to calculate its
height.
horizontal
vx  16m / s
d x  31.0
d x
t
31.0
16 
t
t  1.9 s
vx 
vertical (using up as positive)
v yi  12m / s
a y  9.80m / s 2
d y  ?
d y  v yi t  12 a y t 2
 12(1.9)  12 (9.80)(1.9) 2
 5.1m
The ball is above the fence, which is only 3.5 m high.
7. A person is in a moving elevator. He throws a rotten egg horizontally out of
the moving elevator with a velocity of 5.0 m/s. At the time of the throw, the
elevator was 8.7 m above the ground. The rotten egg landed 4.2 m away from
the elevator. What was the velocity of the elevator? Was the elevator moving
up or down?
vertical (using up as positive)
horizontal
vx  5.0m / s
d y  8.7 m
d x  4.2m
a y  9.80m / s 2
v yi  ?
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d x
t
4.2
5.0 
t
t  0.84 s
vx 
d y  v yi t  12 a y t 2
8.7  v yi (0.84)  12 (9.80)(0.84)2
v yi  6.2m / s
The elevator was moving downward at 6.2 m/s.
8. An airplane is in level flight at a velocity of 500. km/h and an altitude of 1500
m when a wheel falls off.
a. What horizontal distance will the wheel travel before it strikes the
ground ?
vertical (using down as positive)
horizontal
vx  500.km / h  139m / s
v yi  0
d x  ?
d  1500m
a y  9.80m / s 2
d y  v yi t  12 a y t 2
1500  0  12 (9.80)t 2
t  17.5s
d x
t
d x
139 
17.5
d x  2400m
vx 
b. What will the wheel's velocity be when it strikes the ground?
d y 
1500 
v yf2  v yi2
2a
v 0
2
yf
2(9.80)
v yf  171m / s
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vx
vy
v
v f  v 2fx  v 2fy
tan  
v yf
vxf
171
139
 51
 (139)2  (171)2

 220m / s
v f  220m / s, 51 from the ground
9. Police agents flying a constant 185 km/h horizontally in a low-flying airplane
wish to drop an explosive onto a master criminal's car travelling 145 km/h (in
the same direction) on a level highway 88.0 m below. At what angle (with the
horizontal) should the car be in their sights when the bomb is released?
horizontal
vertical (using down as positive)
vxpg  185km / h  51.4m / s
v yi  0
vxcg  145km / h  40.3m / s
d y  88.0m
d x  ?
a y  9.80m / s 2
d y  v yi t  12 a y t 2
88.0  0  12 (9.80)t 2
t  4.24s
Looking at the horizontal, we need to know the velocity of the plane with
respect to the car so that we can find out how much farther the bomb must
travel than the car:
vxpg  vxpc  vxcg
vxpc  vxpg  vxcg
 51.4  40.3
 11.1m / s
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d x
t
d x
11.1 
4.24
d x  47.1m
vx 
So the plane should be 47.1 m behind the car when the bomb is released.
d x
d y
tan  
d y
d x
88.0
47.1
  61.8

The bomb should be in their sights at an angle of 61.80 below the horizontal.
10. A basketball player tries to make a half-court jump-shot, releasing the ball at
the height of the basket. Assuming the ball is launched at 51.0o , 14.0 m from
the basket, what speed must the player give the ball?
v
vy
vx
vx  v cos 
v y  v sin 
 v(cos 51.0)
 0.629v
 v(sin 51.0)
 0.777v
horizontal
vertical (using up as positive)
vx  0.629v
v yi  0.777v
d x  14.0m
a y  9.80m / s 2
d y  0
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d x
t
14.0
0.629v 
t
22.3
t
v
Substituting this into the equation for vertical motion,
vx 
d y  v yi t  12 a y t 2
0  0.777vt  12 (9.80)t 2
4.9t 2  0.777vt
 22.3 
4.9 
  0.777v
 v 
v  11.8m / s
11. A teflon hockey puck slides without friction across a table at constant velocity.
When it reaches the end of the table, it flies of and lands on the ground. For
each of the following questions, draw all vectors to scale.
a. Draw the situation above, drawing vectors showing the force on the
puck at two positions while it is on the table and at two more while it is
in the air.
The force of gravity vectors should all be the same length.
b. Draw vectors showing the horizontal and vertical components of the
puck's velocity at the four points.
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The horizontal velocity components should all be the same length; the
vertical velocity component should be larger in the fourth diagram than
in the third.
c. Draw the total velocity vector at the four points.
The total velocity vector should be larger and more vertical in the fourth
diagram than in the third.
12. Suppose an object is thrown at an angle with the horizontal with the same
initial velocity on the moon, where g is one-sixth as large as on Earth. Will the
following quantities change? If so, will they become larger or smaller?
a. vxi and v yi
Since these are the initial velocity components, they will not change (it
is thrown with the same initial velocity)
b. time of flight
Since the acceleration is smaller, it will take longer for the vertical
velocity to change. The object will therefore be in the air for more time.
c.
maximum height
v yf2  v yi2 v yi2

Since d y 
, a smaller acceleration will result in a larger
2a y
2a y
maximum height.
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d. Range
Range is the horizontal distance travelled. Since the object is in the air
for more time and the horizontal velocity is unchanged, the object will
travel further.
13. A sniper on a building is trying to hit a target on the ground. The building is
13.0 m high. The sniper points his rifle at a point 49.5 m away from the
building. If the bullet travels at 135 m/s, how far from the building will the bullet
land?
We must first use the direction he is aiming the gun to calculate the angle.
49.5
13.0
  75.3
tan  
This angle can now be used to calculate the velocity components.
v
vy
vx
v y  v cos 
vx  v sin 
 135(sin 75.3)
 131m / s
 135(cos 75.3)
 34.3m / s
horizontal
vertical (using down as positive)
vx  131m / s
v yi  34.5m / s
d x  ?
a y  9.80m / s 2
d y  13.0m
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d y  v yi t  12 a y t 2
13.0  34.5t  12 (9.80)t 2
0  4.9t 2  34.5t  13.0
t1  0.359 s
t2  7.40 s
Since the time must be positive,
d x
t
d x
131 
0.359
d x  47.0m
vx 
14. Show that the range R of a projectile, which is defined as the horizontal
distance travelled when the final point is at the same level as the initial point,
is given by the equation
v 2 sin 2
R
g
where v is the initial velocity of the projectile and  is the angle with the
horizontal. (Hint: use the trigonometric identity sin 2  2sin  cos )
v
vy
vx
vx  v cos 
vy  v sin 
Since range R is the same as the horizontal displacement,
d x
t
R
v cos  
t
R
t
v cos 
vx 
And since the vertical displacement is zero
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d y  v yi t  12 a y t 2
0  (v sin  )t  12 ( g )t 2
1
2
gt 2  vt sin 
t
2v sin 
g
Equating these gives
R
2v sin 

v cos 
g
2v 2 sin  cos 
R
g

v 2 sin 2
g
15. Using the equation given in question 14 answer the following questions.
a. Assuming that the initial velocity is v , what angle will provide the
maximum range?
The maximum range will be when sin 2 has its maximum value, which
is one:
sin 2  1
2  90
  45
b. What minimum initial velocity must a projectile have to reach a target
90.0 m away?
Since we want the smallest velocity that will have a range of 90.0, this
must be the maximum range for this velocity:
v 2 sin 2
R
g
R
v2
g
v2
9.80
v  29.7 m / s
90.0 
c. A garden hose held near the ground shoots water at a speed of 4.2
m/s. At what angle(s) should the nozzle point in order that the water
land 1.8 m away? Why are there two different angles?
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R
v 2 sin 2
g
(6.5) 2 sin 2
9.80
sin 2  0.46
2  27 or 153
2.0 
  14 or 76
There are two different angles because the range of a projectile is
symmetrical around the angle of 45o (for example, 20o and 70o or 40o
and 50o), which provides the maximum range. Try this with other
angles that add to give 90o.
16. A ball is thrown horizontally from the top of a cliff with initial speed vo . At any
moment, its direction of motion makes an angle of  with the horizontal.
Derive a formula for  as a function of time.
tan  
v yf
vo
Since the only unknown variable here is v yf , we need an expression for this
involving time.
vertical (using down as positive)
ay 
v yi  0
g
ay  g
v yf  v yi
t
v yf  0
t
v yf  gt
ty  t
v yf  ?
Substituting that into our previous equation gives
gt
tan  
vo
 gt 

 vo 
  tan 1 
UNIT 2 2D Motion
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17. Two cricket balls are thrown horizontally from the same height but at different
speeds. The faster ball hits the wicket at a location higher than the slower one
does. Why does the faster ball not fall as far as the slower one? After all, they
travel the same distance and accelerate down at the same rate.
The slower ball is in the air for more time, since it takes longer to get to the
wicket. This gives the ball more time to fall, so it falls a greater distance while
in the air.
18. A hunter is trying to shoot a monkey hanging from a tree. As soon as the
hunter fires, the monkey is going to let go of the tree. Assume that the hunter
is below the monkey so that he must angle the gun upward at an angle to aim
at the monkey. Should the hunter aim directly at, above, or below the monkey
in order to hit him?
Just as in the similar question the previous section, the hunter should aim
directly at the monkey. This can be shown below; this is an advanced problem
and all trainees should not be expected to be able to provide this solution.
Monkey (using up as positive)
d ym  v yi t  12 a y t 2
v yi  0
a  g
 0  12 gt 2
d ym  ?
  12 gt 2
So the monkey’s height at any time is
h  ho  12 gt 2
where ho is the monkey’s original height.
Bullet
tan  
UNIT 2 2D Motion
ho
x
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Page 37 of 136
horizontal
vertical (using up as positive)
vx  v cos 
v yi  v sin 
d x  x
ay   g
d yb  ?
d x
t
x
v cos  
t
x
t
v cos 
vx 
d yb  v yi t  12 a y t 2
 vt sin   12 gt 2
 x 
2
1
 v
 sin   2 gt
v
cos



1
 x tan   2 gt 2
But tan  
ho
x
h 
d yb  x  o   12 gt 2
 x
 ho  12 gt 2
It can be seen that the monkey’s height is the same as the bullet’s vertical
displacement at time t after travelling a horizontal distance x ; since they are
at the same height, the bullet will hit the monkey.
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