2-D HW 3
Q3.8. Reason: Running while throwing a ball increases the distance of the throw because it increases the horizontal component of the
ball’s velocity without changing the time of flight. Relative to the ground, the ball’s horizontal velocity component is increased by an
amount equal to the speed of the runner. Since the velocity of the runner is purely horizontal, the vertical velocity component of the ball is
unchanged and so the time of flight of the ball is unchanged. It is interesting to note that since, relative to the ground, the ball has a greater
horizontal component of velocity, the angle of the ball’s velocity is higher relative to the person throwing the ball than relative to the ground.
However, neither relative to the ground, nor relative to the person throwing the ball will a launch angle of 45˚ give the maximum range. The
angle 45˚ gives the maximum range for a ball launched by someone not moving because it gives the best compromise between having a high
time of flight and a large horizontal component of velocity. But when the person is running, the horizontal component of velocity gets an
advantage so the time of flight can be greater at the expense of the horizontal component of velocity. Thus the best angle relative to the
person is greater than 45˚.
Assess: While the best launch angle is no longer 45, it is still true that running increases the range of the ball.
Q3.21. Reason: (a) The ball is going along the trajectory, so the best choice for the velocity at position 2 is D.
(b) The direction of the acceleration is not related to the direction of the velocity—only the direction of the change in velocity. The ball is in
free fall, and so its acceleration is down, just like the acceleration of all other objects in free fall. The best choice is A. This does mean,
however, that the change in velocity vectors (v ) must also be down.
(c) The ball at position 3 is moving horizontally (the vertical component is zero for an instant), so choice C is best.
(d) The same argument for the acceleration at position 2 applies at position 3; the acceleration is down. So the answer is A.
Assess: Galileo taught us the law of falling bodies: “All bodies in free fall (projectiles) have the same acceleration.” The ball is in free fall at
both position 2 and position 3 so the accelerations are the same, regardless of the fact that the velocities are in different directions.
Q3.23. Reason: The key to projectile motion problems is to realize that the motion in the x-coordinate is independent of the motion in
the y-coordinate. We can solve an equation in one of these directions and use the results in an equation for the other direction. For example,
t is the same for the horizontal and vertical components of the motion.
(a) First find the horizontal component of the velocity, and, realizing it will be constant, find the time to impact.
(vx )i = vi cos  = (89 m/s) cos 40 = 68.2 m/s
time 
distance
300 m

 4.4 s
speed
68.2 m/s
So the correct choice is C.
(b) The vertical component of the initial speed is
(v y )i  vi sin   (89 m/s) sin 40   57.2 m/s
Now that we know t  4.4s and (v y )i  57.2 m/s we can solve for h (which is yf).
yf  yi  (vy )i t 
1
1
g (t )2  0.0 m  (57.2 m/s)(4.4 s)  (9.8 m/s 2 )(4.4 s) 2  157 m  160 m
2
2
So the correct choice is D.
Assess: The answers to both parts seem reasonable; in either case if we had been off by a factor of 10 in either direction we would think the
result not realistic.
P3.29. Prepare: This problem is asking us for a range so we need the horizontal component of velocity, vx , and the time of flight, t.
The time of flight, in turn, depends on the initial vertical component of velocity, (v y ) i and the overall vertical displacement of the rock, y.
Solve:
We first find v x and (v y ) i using Equations 3.25:
vx  vi cos   (25 m/s) cos 30  21.7 m/s
(v y )i  vi sin   (25 m/s) sin 30  12.5 m/s
Using the equation which relates vertical velocity, acceleration and displacement: (vy ) f 2  (vy )i 2  2ay y , we can find the final vertical
component of velocity.
(vy ) f 2  (12.5 m/s)2  2(9.8 m/s2 )(12 m)
This equation has two solutions: (v y ) f  19.8 m/s . We choose the negative solution since the rock is descending when it lands. Now we
can find the time of flight using a y  v y / t :
t  vy / ay  (19.8 m/s  12.5 m/s) / (9.8 m/s 2 )  3.29 s
Finally, the range of the rock is given by x  vx t :
x  vx t  (21.7 m/s)(3.29 s)  71 m
The rock lands 71 m from the castle wall.
Assess: Notice that the mass of the rock was not involved. So if this had been a baseball thrown from a 12 m tower, it would have gone
equally far—about three quarters of a foot ball field.
P3.66. Prepare: We convert the 83 yards to meters and then use the formula for range at a launch angle of 45.
 36 in   1 m 
83 yd  83 yd 

  75.9 m
 1 yd   39.37 in 
Solve:
The record range is 83 yards or 75.9 m. We use an extra significant figure since this datum will be used to get another result. Based on
the formula for range, we can say: v  gR / sin (2 ) . Using 45 for the angle gives the following:
v
9.8 m/s  (75.9 m)  27 m/s
2
The launch speed of this record pass was 27 m/s.
Assess: The launch speed is about 60 mi/hr. This is impressive for a football. We would expect the launch speed to be less than that for a
baseball because of a football’s greater mass.