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Ball B
Ball A
DQ1) Two cannon balls, each with mass 20 kg, are fired from ground level over a flat plateau. Ball A is fired with an initial speed of 250 m/s at an angle of 30 degrees from the horizontal, and Ball B is fired from the same spot with an initial speed of 200 m/s at an angle of 45 degrees from the horizontal.
a) Which ball spends more time in the air? Check with your group members to see if you all agree and how many ways there are to figure out the answer to this question.
The time in the air is determined by the y component of the motion. Since each ball has the same acceleration (g=9.81 m/s2 downward), the ball with the largest upward velocity will spend the most time in the air. You could also make this conclusion from the velocity-time kinematic equation:
Since these cannon balls land at the same height they were launched from, the total time for these trajectories is twice the time to the maximum height.
The upshot of all of this is that the cannon ball with the largest upward velocity spends the most time in the air. Consequently, all we have to calculate is the y-component of the launch velocities.
Ball A Ball B
vA0,y=(250 m/s) sin(30^0 )=125 m/s 450
vB0,y=(200 m/s) sin(45^0 )=141 m/s
Since B has the largest upward velocity component, Ball B spends the most time in the air.
b) Which ball has the greatest horizontal speed when it lands?
The horizontal speed is unchanged because there is no x-acceleration. Since the final horizontal speed equals the initial, we can just compare the initial x-components of their velocities. The ball with the largest initial x-component of velocity will have the largest final horizontal speed.
Ball A Ball B
vA0,x=(250 m/s) cos(30^0 )=217 m/s
vB0,x=(250 m/s) cos(45^0 )=141 m/s
Since Ball A has the largest initial x-component of velocity, Ball A lands with the largest horizontal speed.
Ball B
c) Which ball has the greatest vertical speed when it lands?
From the symmetry of the problem (i.e. landing at the same height they were launched at), the vertical component of the final velocity will be the opposite of the vertical component of the initial velocity. Using the calculations from part a), we see that the two vertical velocities will be
vAf =-vA0 = -125 m/s
vBf = -vB0 = -141 m/s
Ball B therefore has the greatest vertical speed when it lands.
d) Which ball travels farther horizontally through the air?
Assuming you didnāt calculate the flight time in part a), we need to calculate the flight time for each ball from the vertical motion. Then we can insert that time into the position-time equation for the horizontal motion.
Start by analyzing the motion in the y-dimension to determine the flight time. The ball starts and lands at height zero and has acceleration (-g) (assuming up is positive).
The t=0 solution represents the starting situation. Weāre interested in the later time when the ball hits the ground.
y-dimension
1 2 2 1 2 (^0 )
0 2
0 2 2
0 0 0
0
0
0
sin
sin
sin
sin
y y v yt a ty
v t g t
gt
t v
gt
v
v
t
g
= ļ£«^ ā ļ£¶
We can plug this time into the displacement vs. time for the x-direction to find how far each ball went.
The x-component of the acceleration is zero.
Here we plug in the flight time found from analyzing the y-direction.
x-dimension
1 2 2 1 2 2
2
0 0 0 0
0 0
0
cos cos
sin cos
sin cos
x x v x a tx
x v t t
x v t
v
x v
g
v
x
g
Now we can plug in the numbers for each ball.
travels the farthest horizontal distance.
Ball A
2 2 250 30 30 5517 9 81 (^2)
m s A (^) m s
sin cos m .
x = ^ =
Ball B
2 2 200 45 45 4077 9 81 (^2)
m s B (^) m s
sin cos m .
Ball A x = ^ =
b) At this launch speed (also known as muzzle speed), with what velocity (speed and direction) will the shell strike the target?
Now that we know the initial launch velocity of the cannonball, we can find the time of flight from the motion in the x-direction. We know the x-velocity remains constant. We can find the y-velocity at impact from the velocity-time equation in the y-direction.
We know the initial velocity, so we can find the time of flight from the motion in the x-direction (see the first box of part a)
0
m s
cos m
. s cos
x
t
v Īø
The x-velocity remains constant, but the y-velocity changes due to the acceleration due to gravity.
( )
( ) (^) ( ) ( )
2
0 0 m m s s 0 0 m m s (^) s m s
cos cos
sin
sin.. s .
x x x
y y y
v v a t v t
v v a t v g t
Īø
Īø
The final velocity magnitude is the vector sum of the x- and y-component of the final velocity.
We can use the Pythagorean Theorem to find the final speed from the two components.
2 2
2 2 63 m s 46 6. ms 78 4. ms
v = v x +vy
We can use trigonometry to find the direction of the final velocity
(^1) ( )
m s m s
tan.
tan.
. below the horizontal
46 6. m s
63 m s
Īø
46 6. m s
63 m s
DQ3) A projectile is shot out at 45Ā° with respect to the horizon with an initial velocity of Vx = Vy = Vcomp. When
is the earliest time, t, that the velocity vector makes an angle of 30Ā° with respect to the horizontal? Try this problem on your own first, then ask your group about it and check your answers.
a) t = Vcomp / (g sin 30Ā°) b) t = Vcomp (1 - sin30Ā°) / g c) t = Vcomp tan 30Ā° / g d) t = Vcomp (1 - tan 30Ā°) / g e) t = Vcomp / (g tan 30Ā°)
We know that the projectile will undergo an acceleration due to gravity of g downward. This will change the y-component of the projectileās velocity but will not affect its x-component. This means the velocity angle with the horizontal will change. We have three relationships we can use:
the final velocity makes an angle of 30Ā± with the horizontal
the x-component of the final velocity is unchanged
the y-component of the final velocity can be found from the velocity-time kinematic equation
The projectile velocity changes according to the constant acceleration kinematic equations
0 comp
y y y y
v v a t v V g t
( 0 )
0 comp comp
x x x
x
v v a t
V t
v V
The final velocity makes an angle of 30Ā± with the horizontal, giving us a relationship between the x- and y-components of the final velocity. (^ )
comp^30
tan
tan
y x
y
v v
v V
We can insert this relationship for vy into the kinematic equation involving the time.
( )
( )
comp
comp comp
comp comp comp
tan
tan (^) tan
v y V gt
V V gt
V V
t V
g g
(^)
The two balls spend the same amount of time in the air. We can insert the expressions for the flight times found above.
We can simplify the answer by bringing the factors of 2 and g under the radical sign.
1 2
1
1
2 2
1
sin
sin
sin
sin
t t
v h
g g
g h
v
g
g h
g
gh
v
ļ£« ļ£¶ ļ£«^ ļ£¶
ļ£¬ ļ£· =^ ļ£¬ļ£¬^ ļ£·ļ£·
DQ5) (challenge problem) An elevator accelerates upward from rest, accelerating steadily from rest to 6 m/s in 6 seconds. Three seconds into the acceleration, a bolt falls off the bottom of the elevator.
a) What is the distance between the bolt and the bottom of the elevator after the 6 seconds are up?
We know the acceleration of the bolt (gravitational acceleration g downward). If we can find the acceleration of the elevator, we can use the distance-time constant acceleration kinematic equation for the elevator and the bolt separately. We can then take the difference between them to find the separation as a function of time.
First we must find the acceleration of the elevator (subscript E). We know its initial and final velocities, as well as the time the motion takes. The velocity-time kinematic equation is therefore probably most useful.
( 6 ) (^0 )
2
E E0 E m E E0 s m E (^) s s
v v a t v v a t
ā^ ā
We can write the elevatorās vertical position as a function of time using the distance-time kinematic equation.
( ) ( )
1 2 2 1 2 2 1 2 2
E E0 E0 E E E E
y t y v yt a yt
t a t
y t a t
To analyze the boltās motion, itās easiest to reset the clock to zero (new time tā) when the bolt falls off. Both the elevator and the bolt will
have the same initial position ( y 0 )
and initial velocity ( vy0). Only their
accelerations will differ (+1m/s^2 for the elevator, -9.81 m/s^2 for the bolt)
1 2 2 1 2 1 2 2 2
E 0 0 E
B 0 0 B 0 0
y y
y y y
y t y v t a t
y t y v t a t y v t g t
Take the difference in positions. The only terms that survive come from their different accelerations.
( (^ )) ( (^ ) )
( )(^ )
1 2 1 2 2 2 1 2 2
E B 0 0 E 0 0
E
y y y
y
y t y t y v t a t y v t g t
a g t
The bolt falls for 3 seconds after being released, so tā=3 seconds.
( (^ )^ (^ ))(^ )
1 2 2
E B s^ m^2 sm^2
' s
( ') ( '). s
. m
t
y t y t
