HP 35s Manual

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HP 35s General applications - Part 3
Other applications
Practice solving problems
- Application 1: True Heading
- Application 2: Thrown object
- Application 3: Gas Pressure

hp calculators
HP 35s General applications – part 3
hp calculators - 2 - HP 35s General applications – part 3- Version 1.0
General applications
This training aid will illustrate the application of the HP 35s calculator to several problems in other areas. These
examples are far from exhaustive, but do indicate the incredible flexibility of the HP 35s calculator.
Practice solving problems
Application 1: True Heading
Example 1: Before a plane takes off, a preflight plan must be filed indicating the proposed trip. In these plans, the
known data are true course (TC), wind direction (WD), wind velocity (WV) and true air speed (TAS). Find
the true heading of a planned flight, if TC = 80 degrees, WV = 55 mph, TAS = 180 mph, and the WD is
toward 95 degrees.
The true heading = TC – ASIN ( WV x SIN ( WD – TC ) / TAS )
Solution: Place the 35s into degrees mode by pressing 91
In RPN mode: 80Ï55Ï95Ï80Ãk180¯
¸ºiÃ
In algebraic mode: 80úi55¸k95Ã80Õ¯180
Ï
Figure 1
Answer: 75.46 degrees.
Application 2: Thrown object
Example 1: If a ball is thrown straight upward with a velocity of 95 feet per second, what is the velocity and height of the
object after 2 seconds? Use a value of 32.2 feetper second persecond for g.
Velocity = OriginalVelocity – g x time
Height = Original Velocity x time – ½ x g x t2
Solution: In RPN mode: 95Ï32Ë2Ï2¸Ã (Velocity at t = 2 seconds)
95Ï2¸0Ë5Ï
32Ë2¸2º=Ã (Height at t = 2 seconds)
In algebraic mode: 95Ã32Ë2¸2Ï (Velocity at t = 2 seconds)
95¸2Ã0Ë5¸
32Ë2¸º=2Ï (Height at t = 2 seconds)

hp calculators
HP 35s General applications – part 3
hp calculators - 3 - HP 35s General applications – part 3- Version 1.0
Figure 2
Figure 3
Answer: After 2 seconds, the object is at a height of 125.6 feet and traveling at a velocity of 30.6 feet per second.
Application 3: Gas pressure
Example 1: The internalpressure of a tank of gas is 1100 psi at room temperature (298 degrees Kelvin). What is the
internal pressure if the temperature rises by 35 degrees Celsius?
New Pressure = Old Pressure x New Temperature ÷ Old Temperature
Solution: In RPN mode: 1100Ï298Ï35Ù¸298
In algebraic mode: 1100¸4298Ù35Õ¯298Ï
Figure 4
Answer: The new pressure is 1229 psi.