Page 56 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 4] MOTION IN A VERTICAL PLANE 41

To reach a certain height h, a body thrown upward must have the same initial velocity as the final velocity
√
of a body falling from that height, namely, v = 2gh.
(Air resistance is assumed to be negligible in the problems ofthis chapter.)

SOLVED PROBLEM 4.1

A person at the masthead of a sailboat moving at constant velocity in a straight line drops a wrench. The
masthead is 20 m above the boat’s deck and the stern of the boat is 10 m behind the mast. Is there a
minimum velocity the sailboat can have in order that the wrench not land on the deck?
The horizontal component of the wrench’s velocity equals the forward velocity of the boat, so the wrench hits
the deck directly below the point from which it was dropped. Only if the boat were accelerated would it be possible
for the wrench to miss the deck.

SOLVED PROBLEM 4.2
A stone dropped from a bridge strikes the water 2.5 s later. (a) What is its final velocity in meters per
second? (b) How high is the bridge?
2
(a) v = gt = (9.8 m/s )(2.5s) = 24.5 m/s
2
1
1
2
2
(b) h = gt = (9.8 m/s )(2.5s) = 30.6m
2 2
SOLVED PROBLEM 4.3
A ball is dropped from a window 64 ft above the ground. (a) How long does it take the ball to reach the
ground? (b) What is its final velocity?
1
2
(a) Since h = gt ,
2

2h (2)(64 ft) √
2
t = = 2 = 4s = 2s
g 32 ft/s
2
(b) v = gt = (32 ft/s )(2s) = 64 ft/s
SOLVED PROBLEM 4.4
What velocity must a ball have when thrown upward if it is to reach a height of 15 m?

The upward velocity the ball must have is the same as the downward velocity the ball would have if dropped
from that height. Hence
√

2 2 2

v = 2gh = (2)(9.8 m/s )(15 m) = 294 m /s = 17 m/s
SOLVED PROBLEM 4.5

A ball is thrown downward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it
moving 2 s later? (b) How far does it fall in these 2 s?
2
(a) v = v 0 + gt = 6 m/s + (9.8 m/s )(2s) = 6 m/s + 19.6 m/s = 25.6 m/s
1
2
2
1
2
(b) h = v 0 t + gt = (6 m/s)(2s) + (9.8 m/s )(2s) = 12 m + 19.6m = 31.6m
2 2