Problem 18

A students stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0-m above a flat horizontal beach. How long after being released does the stone strike the beach below the cliff? With what speed and angle of impact does it land?

SOLUTION:

 
KNOWNS
UNKNOWN
yo = 0
t
y = -50.0 m
qo = 0
vo = 18.0 m/s
y = yo + (vo sinqo )t – ½ gt2
-50.0 m = 0 + 0 – ½ (9.80 m/s2)t2
t2 = 2(50.0 m)/( 9.80 m/s2)
t = 3.19 s
 
KNOWNS
UNKNOWN
yo = 0
v
y = -50.0 m
qo = 0
vo = 18.0 m/s
t = 3.19 s
To find the magnitude and direction of the velocity means finding both x and y components of the velocity vector:
vx = (vo cos qo) = 18.0 m/s
vy = (vo sin qo) – gt = 0 – (9.80 m/s2)(3.19 s) = -31.26 m/s
|v| = [(18.0)2 + (-31.26)2]1/2 = 36.1 m/s
qo = tan-1 (-31.26/18.0) = -60.10

Problem 20

A place kicker must kick a football from a point 36.0 m from the goal. The ball must clear the crossbar, which is 3.05 m above the ground. When he kicks the ball, it leaves with a speed of 20.0 m/s at an angle of 53.00 to the horizontal.

(a) By how much does the ball clear or fall short of the crossbar?

SOLUTION:

 
KNOWNS
UNKNOWNS
x = 36.0 m
y
qo = 53.00
vo = 20.0 m/s
yo = 0
y = yo + (vo sinqo )t – ½ gt2
But we do not know what t is so we need to find t first:

x = (vo cos qo)t
36.0 m = (20.0 m/s)(cos 53.0)t
t = 2.99 s

y = yo + (vo sinqo )t – ½ gt2

y = 0 + (20.0 m/s)(sin53.0)(2.99) – ½ (9.80 m/s2)(2.99 s)2
y = 47.76 m – 43.81 m = 3.95 m
The ball does clear the bar by 0.90 m.

b) Does the ball approach the cross bar while still rising or falling?

SOLUTION:
To do this, we need to find the time it takes to reach maximum height. If the time it takes to reach maximum height is less than 2.99 s then the ball was rising, otherwise, the ball is falling.
At maximum height, vy = 0, therefore:

vy = (vo sin qo) – gt
0 = (20.0 m/s)(sin 53.0) – (9.80 m/s2)t
t = 1.63 s therefore FALLING.

Problem 23
A projectile is launched with an initial speed of 60 m/s at an angle of 300 above the horizontal. The projectile lands on a hillside 4.0-s later. Neglect air friction.
(a) What is the projectile’s velocity at the highest point of its trajectory?

SOLUTION:

To find the velocity of a projectile, we must find the velocity in the x and y directions, then find the magnitude and direction of the velocity. This particular question wants the velocity at maximum height and we know that at maximum height, the velocity in the y-direction is equal to ZERO.
KNOWNS
UNKNOWNS
t = 4.0 s
v
qo = 300
vo = 60 m/s
vy = 0
vy = 0
vx = (vo cos qo) = (60 m/s)(cos 300) = 52 m/s
|v| = 52 m/s at 0 degrees to the horizontal.
(b) What is the straight-line distance from where the projectile was launched to where it hits?
 
KNOWNS
UNKNOWNS
t = 4.0 s
x
qo = 300
vo = 60 m/s
x = (vo cos qo)t = (60 m/s)(cos 300)(4.0 s) = 208 = 210 m (due to significant figures).
Problem 24
A jet liner moving initially at 300 mph due east enters a region where the wind is blowing at 100 mph in a direction 30.00 north of east. What is the new velocity of the airplane relative to the ground?
SOLUTION:
To an observer on the ground, once the airplane enters the region with the wind, the plane will appear to be moving other than directly east (as shown by the blue vector in the diagram). The blue arrow in the diagram is a representation of what a person on the ground would observe.
Treat this problem like a regular vector problem where you have to find the displacement of the two red arrows in the figure:
x
y
A
300 cos 0 = 300 mph
300 sin 0 = 0
B
100 cos 30 = 86.6 mph
100 sin 30 = 50.0
Resultant
386.6 mph
50.0 mph
|R| = [(386.6 mph)2 + (50.0 mph)2]1/2 = 390 mph
q = tan-1 (50.0/386.6) = 7.370 north of east.

Problem 26

A river flows due east at 1.50 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 10.0 m/s due north relative to the water.

(a) What is the velocity of the boat relative to the shore?

SOLUTION:

In the figure:

v(br) =velocity of the boat relative to the river = 10.0 m/s

v(rs) = velocity of the river relative to the shore = 1.50 m/s
v(bs) = the unknown quantity which is the velocity of the boat relative to the shore.
The simplest way to solve this is using the Pythagorean theorem:
v(bs)2 = v(br)2 + v(rs)2
v(bs) = [(10.0 m/s)2 + (1.50 m/s)2]1/2 = 10.1 m/s

(b) If the river is 300 m wide, how far down the shore has the boat moved by the time it reaches the shore?

SOLUTION:

First find the total time it takes to reach the shore by dividing 300 m by v(bs):

t = (300 m)/(10.1 m/s) = 29.7 s

Then, use this time to find the total distance traveled by multiplying this time by the velocity of the river:

D = vt = (1.50 m/s)(29.7 s) = 44.5 m

Problem 28

The pilot of an aircraft wishes to fly due west in a 50.0-km/h wind blowing southward. If the speed of the aircraft relative to the wind is 200 km/h, in what direction should the aircraft head and with what magnitude?

SOLUTION:

This is similar to the previous problem. If the desired direction is westward, we have a right triangle with the hypotenuse equal to 200-km/h and one leg equal to 50.0 km/h, and we are interested in finding the magnitude and direction of the hypotenuse:

Vpg = [(200)2 - (50.0)2]1/2 = 194 km/h
q = tan-1 (50.0/194) = 14.50 south of west.