CK-12-Physics - Intermediate

http://www.ck12.org Chapter 3. Two-Dimensional Motion

If all that was asked for in this problem was the instantaneous speed of the dart at 0.25 s we would be done.
However, you may be asked for the dart’s instantaneous velocity att= 0. 25 s. In that case, you need to determine
the instantaneous direction of the dart as well (recall that velocity is a vector quantity, and as such, has a magnitude
anda direction).

Finding the instantaneous direction of the velocity of the dart att= 0. 25 s:

Using the tangent function is the most efficient method. The angle that we need to find is somewhat arbitrary;
after all, other than the right angle, the triangle has two perfectly good angles to choose from. The angle usually
preferred, however, is measured fromx−direction. Using the inverse tangent relationship which can be found on
your calculator, we have:

tanθ= 102 .5m/s.9m/s; tan−^1 θ= 12. 7 ◦

If you are asked to represent the vector using the trigonometric definition of angle measurement, than either− 12. 7 ◦
or 347. 3 ◦would do. The final vector result can be stated as( 11 .2m/s,− 12. 7 ◦).

Components of projectile motion

In the problem that follows we imagine the projectile launched from the ground with an angular elevation between 0
and 90 degrees, with an initialx−component of velocity of +30 m/s and an initialy−component of velocity of +40
m/s.

FIGURE 3.25

Author: CK-12 Foundation - Christopher

Auyeung License: CC-BY-NC-SA 3.0

Some typical questions that can be asked in such a situation are:

1. What is the time the projectile takes to reach the highest position above the ground?


2. What is the projectile’s highest position above the ground?


3. What is the velocity of the projectile at its highest position above the ground?


4. What is the range of the projectile?


5. In order to answer the first question, let’s consider what determines the amount of time the projectile remains
   airborne. A velocity of 30 m/s in thex−direction, does not affect the time in the air; no more so than the gun or
   dart’s airborne time was affected by their horizontal motion. The vertical component of motion, however, must
   determine how much time the projectile spends airborne since only the vertical motion is subject to gravity. If we
   imagine just the vertical component of +40 m/s we can quickly estimate (if we make the approximation that -9.8
   is close enough to -10) the amount of time it takes the projectile to reach its highest position (peak); it’s about 4
   seconds, or if we need more accuracy; 940. 8 = 4. 08 s= 4. 1 s. Remember that the projectile loses 9.8 m/s of velocity
   every second it ascends. Thus,


6. Answer:t= 4. 1 s