3.4. Projectile Motion http://www.ck12.org
- Answer: Once again we can usex=vtthat is,y=vi+ 2 vft, as long as we understand thatvrepresents the average
velocity for theascentpart of the projectile’s motion. Since the projectile began with an initial velocity in the
y−direction of 40 m/s and has a final velocity of 0 m/s, at the top of its motion in they−direction, its average
velocity is(^402 +^0 )= 20 m/s. Its highest position above the ground is therefore,( 20 m/s)( 4. 08 ) = 81. 6 = 82 m. - Answer: It is often incorrectly thought that the velocity of the projectile at its peak position is zero. It must
be remembered that until the projectile hits the ground, it is always moving horizontally. Though the vertical
velocity is zero at the top of its motion, the horizontal velocity is still +30 m/s. Therefore, the velocity of the
projectile at its highest position above the ground is +30 m/s (the + indicating the projectile is moving with a
velocity of 30 m/s to the right). - Answer: The range of the projectile is determined by finding the total time the projectile is airborne and
multiplying that time by its speed in the horizontal direction(x=vt). The projectile’s motion is symmetric
in the absence of air friction so the total time for the trip is twice 4.08 s or 8.16 s. The range is, therefore,
( 30 )( 8. 16 ) = 244. 8 m= 240 m.
Of course, instead of using a bit of physical reasoning and simple arithmetic, we could have used more sophisticated
equations to answer all these questions, but why make things more complicated than they have to be?
Before leaving this problem, let’s alter the given information, in order to see another way the problem could have
been stated.
A projectile is launched from the ground at an angle of 53.13 degrees with a speed of 50 m/s.
Find the answers to questions 1-4 above.
Our claim is that both problems are, in fact,identicaland have the same answers. If we’re not given the horizontal
and vertical components of the velocity, it is good policy to find them before trying to solve a problem of this sort.
Thex−component (horizontal) is:vx=vcosθ→50 cos 53. 13 ◦= 30 m/s
They−component (vertical) is:vyi=vsinθ→50 sin 53. 13 ◦= 39. 999 → 40 m/s,
Since we see that the components are identical to the original problem, we can solve the problem the same way.
http://demonstrations.wolfram.com/ThrowingABaseballFromTheOutfieldToHomePlate/
An interesting aside: If at 4.1 s, (the time when the projectile has reached its maximum height) we pretend to erase the
first half of the projectile’s motion and label the projectile at its peak position with a vector (rather an a component
of a vector) pointing to the right with a speed of 30 m/s, the remaining path of the projectile would be similar to the
bullet and dart of the previous section (3.3 “A Special case of Projectile Motion”) and the problem would be solved
in the same manner as both bullet and dart problems.SeeFigurebelow.
FIGURE 3.26
Author: CK-12 Foundation - Christopher
Auyeung License: CC-BY-NC-SA 3.0Let’s take a trip!