Projectile Motion

Projectile Motion - Answers
1. A diver running 4.6 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 1.5 s later.
How high was the cliff and how far from its base did the diver hit the water?
Known:
Initial Velocity
Flight Time
v
t
= 4.6 m/s , 0°
= 1.5 s
Find: Height of Cliff
Distance from Base
dy
dx
= vyi · t + ½ g t2
= vx · t
Height of Cliff
dy
= vyi t + ½ g t2 = 0 + ½ ( – 9.81 m/s2 ) (1.5 s)2 = –11.03625 m = –11 m
X
Y
d
6.9 m
- 11 m
vI
4.6 m/s
0
 The cliff was 11 m high.
vF
4.6 m/s
Distance from Base
a
0
-9.81 m/s2
t
1.5 s
1.5 s
d x = vx · t
= (4.6 m/s) · (1.5 s) dx = 6.9 m
 The diver hit the water 6.9 m from the base of the cliff.
2. A hunter aims directly at a target (on the same level) 320 m away. If the bullet leaves the gun at a speed of 550 m/s, by
how much will it miss the target?
Known:
Target Distance
Initial Velocity
dx
v
= 320 m
= 550 m/s , 0° (same level)
Find: Distance Fallen
dy
= vyi · t + ½ g t2
Need:
t
= dx / v x
Flight Time
t
= dx / vx = 320 m  550 m/s t = 0.5818 s
Vertical Displacement
dy
Flight Time
t = 0.58 s
= vyi · t + ½ g t2 = 0 + ½ ( - 9.81 m/s2 ) (0.5818… s)2
= -1.6604 m
dy = - 1.7 m
 The bullet will miss the target by 1.7 m.
3.
Initial Velocity
Distance Fallen
v
dy
= 16 m/s , 35° (same level)
= – 1.8 m
Find: Distance (x)
dx
= vx t
Need:
Initial Velocities
vx
vy
= v cos  = 16 cos 35° = 13.1064… m/s
= v sin  = 16 sin 35° = 9.1772… m/s
Flight Time
dy = vyi t + ½ g t2
0 = -4.905t2 + 9.1772…t + 1.8
use quadratic formula
b  b2  4ac 9.1772 (9.17722  4(4.905)(1.8)
9.1772 119.537 9.1772210.933



 2.05s
2a
2(4.905)
9.8
9.81
Distance (x)

Y
d
320 m
- 1.7 m
vI
550 m/s
0 m/s
vF
550 m/s
a
0
-9.80 m/s2
t
0.58 s
0.58 s
An athlete throws the shot put with an initial speed of 16 m/s at a 35° angle to the horizontal. The shot leaves the shot putter's
hand at a height of 1.8 m above the ground. Calculate the distance traveled.
Known:

X
dx
or -0.18s
X
Y
d
27 m
- 2.2 m
vI
13.1 m/s
9.18 m/s
vF
10.7 m/s
a
0
-9.80 m/s2
t
2.05 s
2.05 s
= vx t = (13.1064…)(2.05) =26.868 m
The shot put traveled 27 m.
4.
The world record for women’s long jump is 7.52 m. Assuming her horizontal speed is 8.0 m/s as she leaves the ground, how long
was she in the air and how high did she go?
Known: Initial Velocity
Distance Horizontal
Find:
Flight Time
Max Height
5.
vx
dx
= 8.0 m/s, unknown angle
= 7.52 m
t
=
d x 7.52m

 0.94s
vx
m
8.0
s
He was in the air for 0.94 s.
occurs halfway* through flight when vy=0
Time interval from dymax (where v=0) until landing
dy = viyt + ½ gt2 = 0 + ½ (-9.81m/s2)(0.47s)2= 1.0835…m
He went to a height of 1.1 m.
Ymax
d
7.52 m
1.1 m
vI
8.0 m/s
0 m/s
vF
8.0 m/s
a
0
-9.81 m/s2
t
0.94 s
0.47 s
*(ONLY when landing at same height)
A sniper on a building is trying to hit a target on the ground. The building is 11.0 m high. The sniper aims his rifle at a point
19.5 m away from the building in order to hit the target. If the bullet travels at 235 m/s, how far from the building is the target?
Sniper – Angle  = 33.69° below horizontal
Known:
X
Y
Initial Velocity
v
= 235 m/s
d
18.4 m
-11.0 m
Distance Fallen
dy
= -11.0 m
vI
202.0 m/s
-120.1 m/s
vF
202.0 m/s
a
0
-9.80 m/s2
t
0.091 s
0.091 s
 = tan-1(-11.0m/18.5m) = -30.735°
Find: Angle
= v cos  = 235 cos -30.735° =202.0 m/s
= v sin  = 235 sin -33.69° = -120.1 m/s
Initial Velocities
vx
vy
Flight Time
dy = vyi · t + ½ g t2
0 = -4.905t2 – 120.1t + 11
use quadratic formula
Distance (x)
6.
X
dx

b  b  4ac 120.1 120.1  4(4.905)(11) 120.1 14639.83


 0.09125
2a
2(4.905)
9.81
2
2
The target is 18.4 m from the building.
= vx t = (202.0…)(0.09125…) =18.43255… m

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.
Should the hunter aim directly at, above, or below the monkey in order to hit him?
The hunter should shoot at the monkey in order to hit him because the monkey and the bullet fall at the same rate.
7.
A soccer ball is kicked with a speed of 24.0 m/s at an angle of 37° to the horizontal. How much later does it hit the ground?
Ball – Angle 37°
Known:
Find:
Xball
Ymax
Y
Initial Velocity
v
= 24.0 m/s, 37°
d
88.3 m
Distance Fallen
dy
= 0 (same height)
vI
19.2 m/s
+14.4
14.4 m/s
Initial Velocities
vx
vy
= v cos = 24.0 cos 37° = 19.16725…m/s
= v sin = 24.0 sin 37° = 14.4436…m/s
vF
19.2 m/s
0
-14.4 m/s
a
0
t
2.94 s
Time to Max Height
Flight Time
t
vy  vy
f
a
i

0  (14.4436)
 1.47233s
9.81
t = 2 x Time to max height = 2 (1.47233s) = 2.94466…s

He hits the ground 2.9 s after it is kicked.
0m
-9.81 m/s2
1.4723
2.94 s
8.
A baseball is hit at 30.0 m/s at an angle of 53.0° with the horizontal. Immediately, an outfielder runs 4.00 m/s toward the infield and
catches the ball at the same height it was hit. What was the original distance between the batter and the outfielder?
Known:
Find:
Initial Velocity of ball vball = 30.0 m/s, 53.0°
Velocity of player
vplayer = -4.00 m/s (opposite direction of batted ball)
Distance Fallen
dy
= 0 (same height)
Initial Velocities
vx
vy
= v cos = 30 cos 53° = 18.05445069 m/s
= v sin = 30 sin 53° = 23.9590653 m/s
Time interval
dy = vyi t + ½ g t2, but dy=0,
Xball
Y
Xplayer
d
88.3 m
0m
-19.6 m
vI
18.0 m/s
24.0 m/s
-4.00 m/s
vF
18.0 m/s
-24.0 m/s
a
0
-9.80 m/s2
0
t
4.89 s
4.89 s
4.89 s
so t = -2vyI /g = -2 (23.959 m/s) /(-9.80 m/s2)= 4.8896 s
Distance Ball Traveled dball = vball × t = (18.0544 m/s)(4.8896s) = 88.27914 m
Distance Player Ran dplayer = vplayer × t = (-4.00 m/s)(4.8896s) = 19.5584 m
The ball went 88.3m toward a player who ran 19.6m toward the batter, so the original distance between the batter and outfielder
was 107.8m (88.27914m + 19.5584m).
9.
A football is kicked at an angle of 35° with the horizontal with a velocity of 25.0 m/s. The field goal poles are 37.0 m away and are
3.5 m high. Is the field goal good?
Known:
Initial Velocity
v
= 25.0 m/s, 35°
Find:
Height at 31.0m
or
Distance when it is 3.5m high
Initial Velocities
vx
vy
= v cos θ = 25 cos 35° = 20.4788 m/s
= v sin θ = 25 sin 35° = 14.3394 m/s
Time interval
t = dx / vx = 37.0m / 20.4788 m/s = 1.8067 s
Height at 31.0 m
dy = vyi t + ½ g t2 = (14.3394 m/s)(1.8067 s)+ ½ (-9.81)(1.8067 s)2
= 25.907 – 16.011 m = 9.896 m
The field goal was GOOD because the ball was 9.9 m above the ground at the goals
posts 37.0m away and the ball only had to be 3.5 m high. It was 6.4 m above the
crossbar.
X
Y
d
37.0 m
+9.9 m
vI
20.5
14.3 m/s
vF
20.5
a
0
-9.80 m/s2
t
1.81 s
1.81 s
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 51.0°, 8.5 m from the basket, what velocity must the player give the ball?
Basketball – Angle +51°, no net vertical displacement
Known:
Find:
X
Y
d
8.5 m
0
Initial Velocities in terms of time interval
dx = vx t, so t = dx / vx
dy = vyi t + ½ g t2, but dy=0, so t=-2vyI /g
vI
?
?
tx  t y
vF
0
-9.81 m/s2
Distance Fallen
Range
2v yi
dx

vx
g
dx
2v sin 

v cos 
g
v2 
dy
dx
= 0 (same height)
= 8.5 m
but,
vx = v cos
and
vy = v sin
a
t
 dx  g
(8.5m)(9.81m / s 2 )

 85.24687 m 2 / s 2
v =9.232977 m/s
2 sin  cos 
2 sin 51 cos 51
The initial velocity of the ball was 9.2 m/s at 51.0°.
11. A person is in a moving elevator. He throws a rotten egg horizontally out of the moving elevator with a velocity of 4.0 m/s. At the
time of the throw, the elevator was 18.7 m above the ground. The rotten egg landed 12.4 m away from the elevator. What was the
velocity of the elevator? Was the elevator moving up or down?
Known:
Find:
Horizontal Velocity
Horizontal Distance
Distance Fallen
vx
dx
dy
= 4.0 m/s
= 12.4 m
= – 18.7 m
Time interval
t
Initial Vertical Velocity dy
= dx / vx = 12.4 m / 4.0 m/s = 3.10 s
= vyi t + ½ g t2
m
= (12.0363)(1.9408)+ ½ (-9.80)(1.9408)2
d y Height
½ g t 2 at 31.0
(18.7)  ½(9.81)(3.10) 2  28.43705
vyi 
 = 23.360m – 18.457m

 9.1732m / s
t
(3.10)
31.0
= 4.903 m
The elevator was moving up at 9.2 m/s.
X
Y
d
12.4 m
- 18.7 m
vI
4.0
+9.2 m/s
vF
4.0
a
0
-9.81 m/s2
t
3.10 s
3.10 s
12. An airplane is in level flight at a velocity of 450 km/h and an altitude of 1200 m when a wheel falls off. What horizontal distance will
the wheel travel before it strikes the ground and what will the wheel's velocity be when it strikes the ground?
Known:
Initial Velocity
Distance Fallen
v
dy
= 450 km/h = 125 m/s, 0°
= – 1200 m
vyI= 0
Find:
Flight Time
dy
t2
t
= vyi t + ½ g t2 and vyi = 0 so
= 2 dy / g = 2 (–1200 m)/(–9.81 m/s2) = 244.648318 s2
= 15.64123… s
Horizontal Distance
dx
= vx t = (125.0 m/s)( 15.64123 s)= 1955.1547… m
Final Velocity Down
X
Y
d
1960 m
– 1200 m
vI
125 m/s
0
vF
125 m/s
- 77 m/s
vy2 = vi2 + 2 ay dy = 0 + 2( –9.81 m/s2)(–1200 m) = 5886 m2/s2
vy = ± 76.72 m/s
a
0
-9.81 m/s2
Final Speed
v2
v
= vx2 + vy2 = (125)2 + (76.72)2 = 21511
= 146.666 m/s = 528 km/h
t
15.6 s
15.6 s
Angle

= tan–1 (vyF/vxF) = tan–1 (-76.26)/(125)) = –31.3865°
The wheel travels horizontally 1960 m (1.96 km) before striking the ground at 528 km/h at an angle of 31° to the horizontal.
13. What minimum initial velocity must a projectile have to reach a target 20.0 m away?
Known:
Find:
Horizontal Distance dx
Launch Angle
θ
Vertical Displacement dy
= 20.0 m
= 45°
to get maximum range
=0
same height that it was launched
d
Initial Velocities in terms of time interval
dx = vx t, so t = dx / vx
dy = vyi t + ½ g t2, but dy=0, so t=-2vyI /g
vI
tx  t y
vF
2v yi
dx

vx
g
but,
vx = v cos θ
and
vy = v sin θ
dx
2v sin 

v cos 
g
v2 
 dx  g
(20.0m)(9.81m / s 2 )

 196.2m 2 / s 2
2 sin  cos 
2 sin 45 cos 45
a
X
Y
20.0 m
0
0
-9.81 m/s2
t
v =14.0 m/s
The minimum initial velocity of the projectile must be 29.7 m/s; the projectile must be launched at an angle of 45° to the horizontal.
14. Police agents flying a constant 200.0 km/h horizontally in a low-flying airplane wish to drop an explosive onto a master criminal's
car traveling 130 km/h (in the same direction) on a level highway 78.0 m below. At what angle (with the horizontal) should the car
be in their sights when the bomb is released?
Known:
Find:
Initial Velocity Bomb
Distance Fallen
Car Velocity
vbomb = 200.0 km/h = 55.555… m/s, 0°
dy = – 78.0 m
vcar = 130 km/h = 36.111…m/s, 0°
Relative Horizontal Velocity of bomb (relative to car)
vyI= 0
Y
d
77.6 m
- 78.0 m
vI
19.4
0
vF
19.4
39.1
a
0
-9.80 m/s2
t
3.99
3.99
vbomb = 19.444…m/s
Need ratio of distances
Flight Time
Xbomb
(relative)
dy
t2
t
= vyi t + ½ g t2 and vyi = 0 so
= 2 dy / g = 2 (–78)/(–9.8) = 15.91836735s2
= 3.98978287 s
Relative Distance
dx
= vx t = (19.4444)(3.989…) = 77.57911134 m
Angle
θ
= tan–1 (dy/dx) = tan–1 (-78.0m/77.57911134m) = –45.155°
The bomb should be released at an angle of 45.2° below the horizontal.

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