**Projectile Motion**

**Essential equations.**

v = v0 + at.

x-x0 = v0t + 1/2at^2. Use when time is involved.

v^2 = v0^2 + 2a(x-x0). Use when time is not involved.

n_x indicates the x component of n.

n_y indicates the y component of n.

an ending 0 in a term represents "initial." Example v0 = initial velocity.

max_d = v0(sin(2theta))/g

It's important to understand that in projectile motion, if I were to drop a ball off a ledge and push a ball off a ledge from the same height at the same time, that they would fall to the earth at the same time.

**
1.** The difference between a vector and a scalar is that a vector has
magnitude as well as direction. A scalar is just the magnitude without a direction,
example mass time, temp.

6.

7.

½*g*t^2.

8.

**Problem 1:** A chicken is kicked at ground level with a speed
of 50.0 m/s at an angle of 35 degrees to the horizontal. How much later does
it hit the ground? What is the maximum altitude the chicken will reach? How
far will the chicken end up from initial kicking position? Assume negligible
air resistance.

To determine how much later the chicken will hit the ground, we first break
the initial velocity vector into its components.

50.0m/s sin 35 = V_y = 28.67 m/s

50.0m/s cos 35 = V_x = 40.9 m/s

Thus we use the equation y-y0 = v0t + 1/2at^2. => y-y0 = v0t + 1/2*g*t^2.

0 = V_y*t + 1/2*-9.8*t^2.

0 = -4.9t^2 + 28.67t. Find the roots of the polynomial with the quadratic formula.

*t = 5.85s or 0s.*

To determine the maximum altitude the chicken will reach, we take the time
it takes to travel up, or the time to travel down.

t = V_y/g. t = 27.67/9.8. t = 2.82s. (or in otherwords, just half the elapsed
time)

y-y0 = v0t+ 1/2*g*t^2 once again. y = 28.67*(2.82) + 1/2*-9.8*(2.82^2).

y = 41.88m

Alternatively, you could've solved this with v^2=v0^2+2a(y-y0).

0 = 28.67^2 + 2(-9.8)(y).

y = 41.88m.

To determine how far the chicken will end up from initial kicking position,
we use the horizontal component of velocity and the total elapsed time.

D _x = (V_x)t. d = 40.9*5.85. d = a whopping 239.265m.

**Problem 2:** A duck piloting a bomber was ordered to deliver
a "surprise" to an unsuspecting colony of chickens. If the bomber
is traveling horizontally at 120m/s, 235m up in the sky, how far in advance
must the present be released in order for the chickens to receive it? (Neglecting
air resistance). If the duck couldn't wait and released the present 500m before
the target, at what vertical velocity must it be thrown at to arrive safely
for the chickens? How quickly will this present be traveling when it hits the chickens?

Determine the time it takes for the present to fall 235m: y-y0= v0t + 1/2*g*t^2.

-235m = 1/2*-9.8*t^2. t = 6.9s.

To find how far in advance, use D_x = V_x*t.

D_x = 120*6.9.

* D_x = 828m in advance.*

Now, if the duck released the package 500m before the target, we then find the
time it takes for the package to travel 500m. t= D_x/V_x.

t = 500/120.

*t = 4.17s.*

Then we use: y-y0=v0t + 1/2*g*t^2.

-235m = 4.17v + 1/2*-9.8*(4.17^2).

-149.8m = 4.17v.

*v = -35.9m/s. Thus, the present must be thrown downwards at 35.9m/s*

The present's velocity can be determined by: v^2=v0^2 + 2g(y-y0)

v^2 = -35.9^2 + 2(-9.8)(-235).

*v = 76.78m/s. *

Or, v = v0 + at.

v = 35.9 + 9.8(4.17)

* v = 76.78m/s*

**Problem 3**(Asked by fleet0011): A hot air baloon has just lifted off and is rising at the constant rate of 2.0m/s. Suddenly, one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 10.0 m/s. If the passenger is 2.5m above her friend when the camera is tossed, how high is she when the camera reaches her?

d-d0 = vt.

d-d0 = v0t + 1/2at^2.

d-2.5 = 2t, d = 10t -4.9t^2, 2t+2.5 = 10t -4.9t^2

-4.9t^2 +8t -2.5 = 0

t = 1.2 or .42

d-d0 = vt,

d-2.5 = 2*.42

*d = 3.34m
*

NOTE: Go with the smaller time, 0.42s, because that is the first time the camera and balloon meet.

**Problem 4**(Asked by glad-ko): A gun shoots bullets that leave the muzzle at 249 m/s. If a bullet is to hit a target 159.8 m away at the level of the muzzle, the gun must be aimed at a point above the target. (Neglect air resistance.) a) How far above the target is this point if the angle the gun makes is less than 45°?

Let Vx = v0 cos theta, Vy = v0 sin theta.

The time it takes to reach the target is given by
d/Vx = t

Now, considering the vertical component..

y-y0 = v0t + 1/2at^2

y-y0 = Vy(d/Vx) + 1/2a(d/Vx)^2. same gun and target are same level, so 0 = Vy(d/Vx) + 1/2a(d/Vx)^2.

Factoring, 0 = (d/Vx)(Vy + 1/2a(d/Vx)).

Now we use zero product property and start to solve for x. 0 = 249sin(x) - 4.9*159.8/249cos(x).

maple input> solve(0 = 249*sin(x) - 4.9*159.8/(249*cos(x)),x);

-3.128962159, -1.583426821, 1.558165832, 0.01263049439 are the angles within 0 to 2pi.

We toss out the negatives and 1.558165832. This leaves us with 0.01263049439radians as our angle, call that theta.

Now, tan(theta) = height above target/159.8m.

height above target ~= *2.018m*.

**Problem 5**(Asked by fleet0011): A basketball is thrown horizontally with an initial speed of 4.20 m/s. A straight line is drawn from the release point to the landing point makes an angle of 30.0 deg. with the horizontal. What was the release hight?

draw a triangle. tan 30 = opp/adjacent. adjacent = d = 4.20*t. opposite = h = -1/2*9.8*t^2.

solve for t, -2*Sqrt(3)/7s. substitute back into h = -1/2*9.8*t^2

h = -1.2m

**Problem 6**(asked by elf222? I forgot):A test rocket is fired vertically upward from a well.
A catapult gives it an initial velocity of 76.9 m/s at ground level, Subsequently, its engines
fire and it accelerates upward at 4.13 m/s^2 until it reaches an altitude of 887m. At that point
its engines fail, and the rocket goes into free fall with an acceleration of -9.80 m/s^2.

(a) How long is the rocket in motion above ground in seconds?

a) y-y0 = v0t + 1/2at^2.

use the a representing net acceleration..

887 = 76.9*t + .5*4.13*t^2.

t = ...

now with t, v2 = v0 + at, find the initial velocity v2 for the free fall.

then do y-y0 = v0t + 1/2at^2 again with a as -9.8, v2 as v0, y as 0 and y0 as 887

(b) What is its maximum altitude?

use v0 from part a and do v^2 = v0^2 + 2ad with v as 0, a as -9.8, solve for d. then add that to 887

(c) What is its velocity just before it collides with the Earth?

with the max altitude from b), do v^2 = v0^2 + 2ad, solve for v with d as the total displacement from
max to earth and a as -9.8 and v0 as 0, since it's at max height

**Problem 7**(forgot who asked this one too, doh!):
A ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 m during
the second 5 s of its motion. How far did it roll during the first 5 s of motion?

x-x0 = v0t + 1/2at^2.

let v2 = initial velocity for second part .

150 = v2*5 + .5*a*25

v2 = v0 + at, started at rest, v0 = 0. v2 = at. v2 = 5a, substitute and solve for a.

then do x-x0 = v0t + 1/2at^2 with v0 as 0, t as 5 and x-x0 as what you're solving for.

**Problem 8**(asked by afterburn): The distance from the pitcher's mound to home plate is 18.4m, the mound is .2m above the level
of the field. A picher throws a fastball with an initial speed of 37.5m/s at the moment the ball leaves the
pichers hand, it is 2.3 meters above the mound. What should the angle between the velocity and the horizontal
be so that the ball crosses the plate .7m above the ground?

dy = 0.7m - (0.2m + 2.3m)

dy = -1.8m.

in the vertical, we have y = v0t + 1/2at^2.

Making the substitutions, we have -1.8 = 37.5sin(theta)t -4.9t^2.

Now lets look at the horizontal. We know that it is independent of the vertical and undergoes no acceleration.

We can then say x-x0 = v0t.

v0 in this case will be the horizontal component of velocity, 37.5cos(theta)

the horizontal displacement is given as 18.4m

So we have 18.4 = 37.5cos(theta)*t.

solving for t.. 0.490667/cos(theta) = t.

Referring back to our equation for the vertical.. -1.8 = 37.5(sin(theta)t - 4.9t^2..

We can substitute t in.

-1.8 = 18.4tan(theta) - 1.17969/cos^2(theta).

With some algebra trickery, we arrive at theta = -1.926

**Problem 9**: An example of the Coriolis effect. Suppose air resistance is
negligible for a golf ball. A golfer tees off from a location
precisely at theta_i = 35.0° north latitude. He hits the
ball due south, with range 285 m. The ball’s initial velocity
is at 48.0° above the horizontal. (a) For what
length of time is the ball in flight? The cup is due south
of the golfer’s location, and he would have a hole-in-one
if the Earth were not rotating. The Earth’s rotation makes the tee move in a circle
of radius R_earth cos theta_i = (6.37*10^6 m) cos 35.0°, completing
one revolution each day. (b) Find the eastward
speed of the tee, relative to the stars. The hole is also
moving eastward, but it is 285 m farther south and thus
at a slightly lower latitude theta_f . Because the hole moves
eastward in a slightly larger circle, its speed must be greater than that of the tee. (c) By how much does the
hole’s speed exceed that of the tee?

a) Using kinematics, we can do y-y0 = v_y*t + 1/2at^2 in the y and x-x0 = v_x*t in the x. Solving the equations 0 = vsin(48)t-4.9t^2 and 285=vcos(48)t for t, t = 8.037s.

b) To find the eastward speed, we use v = rw along with R_e cos theta_i = r. w = 1 rev/day * 1day/24hr * 1h/60m * 1m/60s * 2pi rads/1 rev = pi/43200 rad/s for omega. v = 379.46356m/s

c) We can find theta_f by finding the angle rotated through in part b. Lets call the angle due to part b theta_2 and then we have theta_f = theta_i - theta_2. since theta_2 = l/r, and vt = l, we then have theta_2 = 0.00256 degrees and theta_f = 34.9974 degrees.

Now we can find the excess speed v. Let's call the hole's speed v_hole and the speed of the tee v_tee. Using v = rw again, v_hole = R_earth cos theta_i * pi/43200. v_tee = R_earth cos theta_2 * pi/43200. v is the difference, R_earth*pi/43200(cos theta_f - cos theta_i). v is now 0.012m/s.

**Problem 10**: A ball is dropped from rest at a height h above ground. Another ball, beneath the first, is launched vertically at the first ball. This happens simultaneously. Find the speed of the second ball if they are to collide at height h/2.

For the first ball

y-y0 = -1/2gt^2

For the second ball

y-y0 = v0t - 1/2gt^2

We want them to collide at a height h/2, so their final position "y" must be h/2.

h/2 - h = -1/2gt^2

h/2 = v0t - 1/2gt^2

Solving for time in the first equation, we have Sqrt[h/g]=t

Substituting that into the second equation..

h/2 = v0*Sqrt[h/g]-1/2h

This leaves us with v0 = h/Sqrt[h/g].

**Problem 11: **A projectile is fired up an incline (incline angle phi) with an initial speed v0 at an angle theta with respect to the horizontal, theta > phi. Show that the projectile travels a distance d up the lincine, where d = 2v0^2cos(theta)sin(theta-phi)/(g*cos^2(phi))

This sounds like a beast of a problem, but can be broken down and manipulated through trig identities into the desired form.

Since d is along the incline, this could be approached by finding the x component and then the y component, then resolving.

The x-component of the trajectory will be x = v0cos(theta)t.

The y-component of the trajectory will be y = v0sin(theta)t - 1/2gt^2.

Now looking at the impact point on the incline..

x = dcos(phi)

y = dsin(phi)

Now that these have been determined, we can begin removing the time variable.

x/(v0cos(theta)) = t

y = xsin(theta)/cos(theta) - 1/2gx^2/(v0cos(theta))^2

great, now we have an expression for the y component with time removed.