Projectiles

In beginning physics, we always neglect air drag because we are not smart enough to handle it yet. So keep in mind that what we say here is very accurate for a baseball given a little toss, but  inaccurate for a ninety mph fastball.

Neglecting air drag, there is only gravity to consider, and it is an experimental fact that it causes the same acceleration as in the free-fall case: g straight down. Thus v sub x (v_x ) does not change, and v_y changes the same way as a free faller. So any projectile motion problem is two problems in one: the x direction where the only equation you will ever use is x = v_xt, and the y direction where you may use any equation from the set of linear motion equations. Just make sure you use only y components in your y direction equations: y, v_y and a_y. And if you make the up direction positive, then a_y=-g

y direction
y = ¹/₂(v_yo+ v_y)t
y = v_yot - ¹/₂gt²
(You finish the list. Go to linear motion if you need to.)

Please understand that a projectile which has a particular velocity at some instant has its future path determined by that velocity. It doesn't matter if it was thrown, kicked, batted, shot from a bow or spat from a cobra. After it is no longer in contact with the thing that pushed it, there is no force from that thing. There is only gravity acting.

Be very careful about signs. Choose a positive direction for y (up or down), and use that also for the y component of velocity and acceleration. If you choose up positive, then a_y = -g.

OTHER HINTS:

1. Think about the highest point in a projectile's path. If it is not obvious to you what v_y is at this point, you need to go back into components of vectors. Talk it over with a friend. This is serious: you may need to enter a 12-step program on vectors. It probably won't help to just look at vectors.

2. Imagine playing catch with a friend, and assume the ball is caught at the same elevation as it had when it left the thrower's hand. There is a symmetry in its path. That is, if you sketch the path on paper, you could fold the paper on a vertical line through the highest point, and the left side would match the right side. Furthermore, the time to get to the highest point equals the time to come back down. These ideas are handy only for the case of the initial position at the same height as the final.

3. Of the quantities x, v_x and t, if you are given two of them, calculate the 3^rd from
x = v_xt. This is simply a consequence of the fact that the horizontal component of v is a constant, because the acceleration is in the vertical direction and hence only v_y changes.

4. To repeat what I said about the y direction: y, v_y a_y and t are related by the linear motion equations. Gravity causes the a_y and it is the free-fall acceleration in the downward direction.

Here is what a projectile's locations look like at equal time intervals if air drag can be neglected.

SAMPLE PROBLEM: When you take your last second jump shot with the basketball game on the line, the ball leaves your hand with v = 20 ft/s at an angle of 60^o above horizontal and the center of the ball is a height 7.0 ft above the floor. The basket is 10 feet above the floor. If the ball makes it into the basket without hitting the rim, what is the horizontal distance to the center of the basket? Do it before reading on.

Really do it first, then read below. You won't get this very well by reading solutions unless you have tried it. Did you learn how to ride a bicycle by watching other people ride?

The components of v: v_x = 20cos60^o = 10 ft/s; v_yo = 20sin60^o = 17.32 ft/s. (Round off later.) We need the time of flight t, then we can find x = v_xt. Use
y = v_yot + ¹/₂a_yt² to find t. Taking the initial location as the origin, y at the basket is
3 ft. Using the quadratic formula, show that

t = [-v_yo ± (v_yo²+ 2a_yy)^1/2]/(a_y).

(Do not use the above equation without deriving it from basic principles.)
Plug in v_yo= 17.32, a_y= -32, and y = 3, and get t = 0.2165 s and 0.866 s. There are two times because it reaches that height on the way up then on the way down. So choose the latter (later) time. Check the units: ft/s divided by ft/s² is s. Plug in:
  x = v_xt = 8.7 ft.

If you want to get the maximum horizontal distance for a projectile, if the initial and final elevations are the same, use 45^o . If the final is above the initial, maximum x is achieved with an initial angle greater than 45^o; If the final elevation is below the initial, use less than 45^o. Calculate the final speed from v_o and y, then use tanq = v_o/v_f. This will give you a maximum x of
x_max = v_ov_f /g. Go here for more on this. For a derivation of the general equation for x,
x = (v_ov_f/g)sin(q+f), go here.

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