Page 57 - Schaum's Outline of Theory and Problems of Applied Physics

42 MOTION IN A VERTICAL PLANE [CHAP. 4

SOLVED PROBLEM 4.6
A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving
1 s later? In what direction? (b) How fast is it moving 2 s later? In what direction?
2
2
(a) We consider upward as + and downward as −. Then v 0 =+6 m/s and g =−9.8 m/s , so
2

v = v 0 + gt = 6 m/s + (−9.8 m/s ) 1 s = 6 m/s − 4.9 m/s = 1.1 m/s
2
which is positive and hence upward.
(b) After 2 s

2
v = v 0 + gt = 6 m/s + (−9.8 m/s )(2s) = 6 m/s − 19.6 m/s =−13.6 m/s
which is negative and hence downward.

SOLVED PROBLEM 4.7
A person in a closed elevator with no floor indicator does not know whether the elevator is stationary,
moving upward at constant velocity, or moving downward at constant velocity. To try to find out, the
person drops a coin from a height of 2 m and times its fall with a stopwatch. What would be noted in
each case?
Since the coin has exactly the same initial velocity when it is dropped as the elevator, this experiment would
give the same time of fall in each case. This time of fall is

2h (2)(2m) √
2
t = = = 0.408 s = 0.64 s
g 9.8 m/s 2
However, if the elevator were accelerated upward or downward, the time of fall would be, respectively, less or more
than this.

PROJECTILE MOTION
The formulas for straight-line motion can be used to analyze the horizontal and vertical aspects of a projectile’s
flight separately because these are independent of each other. If air resistance is neglected, the horizontal velocity
component v x remains constant during the flight. The effect of gravity on the vertical component v y is to provide
a downward acceleration. If v y is initially upward, v y first decreases to 0 and then increases in the downward
direction.
The range of a projectile launched at an angle θ above the horizontal with initial velocity v 0 is
v 2 0
R = sin 2θ
g
The time of flight is
2v 0 sin θ
T =
g
If θ 1 is an angle other than 45 that corresponds to a range R, then a second angle θ 2 for the same range is
◦
given by
◦
θ 2 = 90 − θ 1
as shown in Fig. 4-1.