# Worked Problems

Find the sum of the following displacement vectors:

 = 5.0  m at 37 o N of E = 6.0  m at 45 o N of W = 4.0  m at 30 o S of W = 3.0  m at 60 o S of E

Solution:

a)
Resolve each vector into components:
 Ax = 5cos 37 o = 3.99  m Ay = 5sin 37 o = 3.01  m Bx = - 6cos 45 o = - 4.24  m By = 6sin 45 o = 4.24  m Cx = - 4cos 30 o = - 3.46  m Cy = - 4sin 30 o = - 2.00  m Dx = 3cos 60 o = 1.50  m Dy = - 3sin 60 o = - 2.60  m

b)
Add up all the x-components and all the y-components:
 Rx = Ax + Bx + Cx + Dx = - 2.21  m Ry = Ay + By + Cy + Dy = 2.65  m

c)
Find || and :
 R = = 3.45  m = = 50 o N of W

The current in a river is 1.0   m/s. A woman swims 300 m downstream and then back to her starting point without stopping. If she can swim 2.0 m/s in still water, find the time of the round trip.

Solution:
We need to find the velocity of the woman relative to the shore for each part of the swim. Let downstream be the positive direction and let vw be the velocity of the water. vw/w is the velocity of the woman relative to the water and vw/s is the velocity of the woman relative to the shore. Then: (i) going downstream vw/s = vw + vw/w = 1.0 + 2.0 = 3.0  m/s (ii) going upstream vw/s = 1.0 - 2.0 = - 1.0  m/s .

To find the time to go 300 m in each direction use x = v0t + at 2. With a = 0 we have t = x/v0 .

This gives (i) downstream:

 td = = 100  s
(ii) upstream:
 tu = = 300  s
The total time of the swim is tt = 100  s + 300  s = 400  s .

The woman in the previous problem swims accross the river to the opposite bank and back. The river is 300 m wide and she swims perpendicular to the current so she ends up downstream from where she started. Find the time for the return trip.

Solution:
Since the woman swims perpendicular to the current we can define the y-axis as parallel to the river and treat the x and y motion independently. We are only interested in the motion in the x-direction. For the first half of her swim we have:

 ax = vx0 = vx = 2.0  m/s x = 300  m.
To find the time to cross the river we use x = vx0t + axt 2 which gives,

 t = = = 150  s. (3)

Since the motion is symetric, the time to return is the same as the time to cross. The total time is tt = 2(150  s ) = 300  s .

(Example 3.4) A plane drops a package of emergency rations to a stranded party of explorers. The plane is travelling horizontally at 40.0 m/s at 100 m above the ground. Find a) where the package strikes the ground relative to the spot it was dropped and b) the velocity of the package just before it hits the ground.

Solution:
Set up the coordinate system as in the Figure 3.5. Consider the x and y components separately. We are given:

 x-motion y-motion x =? y = - 100  m v0x = 40  m/s v0y = 0 ax = 0 ay = - 9.8 m/s 2

a)
First we find the time of flight from the y-motion.
 y = voyt + ayt 2 - 100 = (- 9.8)t 2 t = 4.5  s. (4)
Then we can find x from,
 x = vx0t + axt 2 x = 40(4.5) + 0 = 180  m. (5)
b)
We find vy from,
 vy = vy0 + at = 0 - 9.8(4.5) = - 44.1  m/s.
Note that the negative sign correctly indicates that the package falls downward. Since ax = 0 , we have vx = v0x = 40  m/s . We can combine the two velocity components to obtain,
 v = = 59.5  m/s = = 48 o . (6)

A projectile is fired with an initial speed of 113 m/s at an angle of 60 o above the horizontal from the top of a cliff 49 m high (see Figure 3.6). Find a) the time to reach the maximum height, b) the maximum height, c) the total time in the air, d) the horizontal range and e) the components of the final velocity just before the projectile hits the ground.

Solution:
Set up the coordinate system.

Consider the x- and y-motion separately. We are given:

 x-motion y-motion x =? yB = - 49  m v0x = 113cos 60 o v0y = 113sin 60 o v0x = vAx = vBx vAy = 0 ax = 0 ay = - 9.8 m/s 2

a)
Find the time to reach the maximum height:
 vAy = v0y + aytA tA = = = 9.99  s. (7)
b)
Find the maximum height:
 yA = v0ytA + aytA2 = 113sin 60(9.99) - (9.8)(9.99)2 = 489  m. (8)
c)
Find the total time tB in the air:
 yB = v0ytB + aytB2 - 49  m = (113sin 60)tB - (9.8)t 2B 0 = 4.9tB2 - 97.9tB - 49. (9)
 tB = (97.9 )
which gives, tB = 20.5  s or tB = - 0.49  s. We reject the second solution (it gives the time the projectile would have left the ground, if it had been thrown from there).
d)
Find the horizontal range ( xB ):
 xB = v0xtB + axtB2 = (113cos 60)(20.5) = 1158  m. (10)
e)
Find the components of the final velocity ( vBx , vBy ):
 vBx = v0x = 113cos 60 = 56.5  m/s vBy = v0y + aytB = 113sin 60 - 9.8(20.5) = - 103  m/s. (11)
Note that the negative value of vBy correctly gives the direction as down.

(Problem 3.50) A projectile is fired at a falling target. The projectile leaves the gun at the same instant that the target falls from rest. Assuming that the gun is initially aimed at the target, show that the bullet will hit the target.

Solution:
Let xB and yB be the x and y positions of the bullet, and let xT and yT be the x and y positions of the target. We need to show that, yB = yT and xB = xT at some common time tc , the time at which the bullet will hit the target.

The motion of the bullet is described by the two equations:

 xB = v0cos   t yB = v0sin   t - gt 2.
It will take the bullet some time, tc , to arrive at the x position of the target, xT . At that time, the bullet and target will be at the same x postion (the target's x postion does not change since it falls straight down). Thus, we have at time tc :
 xT = xB = v0cos   tc
at time tc .

In order for the bullet to hit the target, the y positions must be equal at tc . The y position of the target is given by

 yT = y0 - gtc2
and for the bullet
 yB = v0sin   tc - gtc2.
We can see from the above equations that in order for yB = yT at tc , we need to have y0 = v0sin   tc .

To show this, consider Figure 3.7:

 tan = = y0 = xT.
Using the previous result, xT = v0cos   tc or cos = (xT)/(v0tc) gives:
 y0 = = v0sin   tc.
Thus we have show that at time tc , xT = xB and yT = yB , and therefore that the bullet will hit the target.