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MECHANICS. 25 1
number of seconds consumed in the ascent and descent, and
this multiplied by cos. of the angle a\ therefore,
__ . , /2 z/ sin. a\ 2 v1
sin. a cos. a
Horizontal range = v y
1 cos. a = —
but 2 X sin. a X cos. a is always equal to sin. of an angle of
twice as many degrees as the angle a. Therefore, the formula
. . v2
sin. 2 a
^educes to horizontal range =
Thus, the following formulas will apply to bodies projected
at an angle. ( See Fig. 1).
The greatest possible height will be,
t v2 sin.2
a
~ ^_
The greatest possible range will be,
, v2 sin. 2 a
S
The time in seconds will be,
2 v sin. a v sin. a
g 0.5 £
v = Velocity in feet per second.
g = Acceleration of gravity = 32.2.
to find the height to which a body can ascend.
Rule.
Multiply the velocity in feet per second by the sine of the
angle (to the horizontal line), square this product and divide
by 64.4, and the quotient is the height in feet.
to find the longest possible range.
Rule.
Multiply the square of the velocity in feet per second by
sine of an angle of twice as many degrees as the angle of the
throw (to the horizontal line), and divide by 32.2. The
quotient is the longest distance the body can be thrown.
to find the time of flight.
Rule.
Multiply the velocity in feet per second by sine of the
angle (to the horizontal line), and divide by 1(3.1. The
quotient is the time in seconds.
Example.
A body is projected at an angle of 55° to the’horizontal
line, with an initial velocity of 120 feet per second. How high
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