## Projectile motion

We have studied the kinematic equations for one-dimensional motion with constant acceleration in module 1.

#### Review problem:

A ball is thrown directly downward with an initial speed of 8 m/s from a height of 30 m.   After what time interval does the ball strike the ground?

Solution:
We have motion in one dimension with constant acceleration.  Put the origin of your coordinate system at the ground and let the x-axis point straight up. Then we know that
xi = 30, vxi = -8 m/s, ti = 0,
xf = 0 m, vxf = ?, tf = ?,
ax = -9.8 m/s2.
We are asked to solve for t = tf.
We have xf - xi = vxi∆t + ½ax∆t2,  -30 m = -8 (m/s) t - 4.9 (m/s2) t2.
We rewrite this equation as t2 + (8 s/4.9) t - (30 s2/4.9) = 0,
or t2 + (1.63 s)t - 6.12 s2 = 0.
This is a quadratic equation with two solutions.
t = -(1.63 s)/2 ± sqrt[((1.63 s)/2)2 + 6.12) = -0.816 s ±  2.60 s.
Only the + sign makes sense. We find t = 1.79 s.

We can also use vxf2 = vxi2 + 2ax(xf - xi),
vxf2 = 64 (m/s)2 + 2*9.8*30 (m/s)2.
vxf = -25.53 m/s.  (The ball is moving downward.)
vxf - vxi = ax t.
-17.53 m/s = -9.8 m/s2*t, t = 1.79 s.

For motion in three dimensions with constant acceleration the kinematic equations are
a = (ax, ay, az) = constant,  ∆vx = ax∆t,  ∆vy = ay∆t,   ∆vz = az∆t.
The x-component of the acceleration only changes the x-component of the velocity, the y-component of the acceleration only changes the y-component of the velocity, etc.
The velocity as a function of time is given by

vx = v0x + ∆vx = v0x + ax∆t,
vy = v0y + ay∆t,
vz = v0z∆t + az∆t,
or   v = v0 + a∆t.

Note: If the directions of v0 and a are different, the direction of v is different from the direction of v0.

For constant acceleration the motions along the perpendicular axes of a Cartesian coordinate system are independent and can be analyzed separately.
The position of the particle at time t is given by

x = x0 + v0x∆t + ½ax∆t2,
y = y0 + v0y∆t + ½ay∆t2,
z = z0 + v0z∆t + ½az∆t2,
or    r = r0 + v0∆t + ½a∆t2.

Note: The directions of r0, v0, a, and r may all be different.  If a is constant then the x, y, and z-coordinated as a function of time can be found independently.

Let us define projectile motion as the motion of a particle through a region of three-dimensional space where it is subject to constant acceleration.

For example, an object moving through the air near the surface of the earth is subject to the constant gravitational acceleration g, directed downward.  If no other forces are acting on the object, i.e. if the object does not have a propulsion system and we neglect air resistance, then the motion of the object is projectile motion.  (This is the reason motion with constant acceleration is called projectile motion.)

Assume that we want to describe the motion of such an object, starting at time t = 0.  Let us orient our coordinate system such that one of the axes, say the y-axis, points upward.  ax = az = 0, ay = -g.  We can rotate our coordinate system about the y-axis until the velocity vector of the object at t = 0 lies in the x-y plane, and we can choose the origin of our coordinate system to be at the position of the object at t = 0.

With the appropriate orientation of coordinate axes we can can treat projectile motion as motion in two dimensions with
vx = v0x,  x = x0 + v0xt,  vy= v0y + ayt,  y = y0 + v0yt + ½at2.

If ay = -g, then
vx = v0x,  x = x0 + v0xt,  vy= v0y - gt,  y = y0 + v0yt - ½gt2.

Let ay = -g.  If the initial velocity v0 makes an angle θ0 with the x-axis, then
v0x = v0cosθ0 and v0y = v0sinθ0
vx = v0cosθ0 = constant,  x =  x0 + v0cosθ0t,
vy = v0sinθ0 - gt,  y = y0 + v0sinθ0t - ½gt2.

We can solve x - x0  = v0cosθ0t for t in terms of x - x0, t = (x - x0)/(v0cosθ0) and substitute this expression for t into
y = y0 + v0sinθ0t - ½gt2.  We obtain
y = y0 + (x - x0)tan(θ0) - g(x - x0)2/(2v02cos20)),

an equation for the path or trajectory of the object.  This equation is of the form y = y0+ a(x - x0) - b(x - x0)2, which is the equation of a parabola.  If x0 = y0 = 0 the parabola passes through the origin.

The trajectory for projectile motion is a parabola.

#### Example:

Assume a projectile is launched with x0 = y0 = 0, v0x = 4 m/s, v0y = 3 m/s.
We have tanθ0 = v0y/v0x = 3/4, θ0 = 30.87o, v02 = v0x2 + v0y2, v0 = 5 m/s.
The projectile moves along a parabolic path until it impacts the ground.  Its coordinates as a function of time are
x = (4 m/s)t, y = (3 m/s)t - (4.9 m/s2)t2
Its velocity components are vx = 4 m/s, vy = (3 m/s) - (9.8 m/s2)t2.

As we can see from the graphs to the right, the projectile impacts the ground after approximately 0.6 seconds.  It reaches its maximum height after approximately 0.3 seconds.  Its range is approximately 2.4 meters.  In approximately 0.3 seconds it has covered half its range.  The trajectory for projectile motion is symmetric about the point of maximum height.  The projectile covers the same horizontal distance reaching its maximum height as it does falling from its maximum height back to the ground.  It takes the projectile the same amount of time reaching its maximum height as it does to fall from the maximum height back to the ground.  When the projectile reaches its maximum height, after approximately 0.3 s, the vertical component of its velocity is zero.

We can find the time when a projectile reaches its maximum height by setting vy = vy0 - gt = 0 and solving for t.  We find
tmax_height= vy0/g = v0sinθ0/g.

We can now find the range R by substituting t = 2tmax_height into the equation for x(t).
R = v0cosθ02tmax_height = (2v02cosθ0sinθ0)/g = (v02sin2θ0)/g.

Similarly, we find the maximum height h by substituting tmax_height into the equation for y(t).
hmax = v0sinθ0tmax_height - ½gt2max_height = (v02sin2θ0)/2g.

Look at the expression for the range, R = (v02sin2θ0)/g.  For a given v0, R as a function of the launch angle θ0 has its maximum value when sin2θ0 has its maximum value of 1.  This happens when 2θ0 = 90o, or θ0 = 45o.
Rmax = v02/g.

#### Problem:

One strategy in a snowball fight is to throw a snowball at a high angle over level ground.  While you opponent is watching the first one, you throw a second snowball at a low angle timed to arrive before or at the same time as the first one.  Assume both snowballs are thrown with a speed of 25 m/s.  The first is thrown at an angle of 70o with respect to the horizontal.
(a)  At what angle should the second snowball be thrown to arrive at the same point as the first?
(b)  How many seconds later should the second snowball be thrown after the first to arrive at the same time?

Solution:
(a) The range of a projectile is R = (v02sin2θ0)/g.
If v0 is constant, the range is a function of θ0
The function sin2θ0 is symmetric about 2θ0 = 90o, or θ0 = 45o.
It has the same value for θ0 = 45o + θ' as it has for θ0 =  45o - θ'.
So a snowball thrown with θ0 = 70o = 45o + 25o has the same range as a snowball thrown with θ0 = 45o - 25o = 20o.
The second snowball should be thrown at an angle of 20o.

(b) The time a snowball is in the air is 2tmax_height = 2vy0/g = (2v0sinθ0)/g.
The flight time of the first snowball is 2tmax_height = 2v0sin70o/g = 4.79 s.
The flight time of the second snowball is 2tmax_height = 2v0sin20o/g = 1.74 s.
The second snowball must be thrown 3.05 s after the first ball.

#### Problem:

An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15 m if her initial speed is 3 m/s.  What is the free-fall acceleration on the planet?

Solution:
To have maximum range for a given initial velocity, her launch angle must be θ0 = 45o.
Her range then is R = (v02sin2θ0)/g' = v02sin90o/g' = v02/g'.
We have g' = v02/R = 0.6 m/s2.

#### Problem:

An owl is carrying a mouse to the chicks in its nest.  Its position at t = 0 is 4.0 m west (x0 = -4.0 m) and 12.0 m above the center of the 30.0 cm diameter nest (y0 = 12.0 m).  The owl is flying east at 3.5 m/s at an angle 30o below the horizontal, i.e. vx0 = (3.5 m/s)cos(30o), vy0 = -(3.5 m/s)sin(30o), when it accidentally drops the mouse.
Is the owl lucky enough to have the mouse hit the nest?
Calculate the horizontal position of the mouse when it has fallen 12.0 m.

Solution
Given: x0, y0, vx0 = v0cosθ0, vy0 = v0sinθ0, θ0 = -30o, ay = -g.
Use vx = v0cosθ0 = constant,  x =  x0 + v0cosθ0t,  vy = v0sinθ0 - gt,  y = y0 + v0sinθ0t - ½gt2.
Solve y - y0 = v0sinθ0t - ½gt2 for t, to find the time it takes the mouse to fall 12 m.
(-12 m) = (-3.5 m/s)*0.5*t - 0.5*(9.8 m/s2)*t2
t2 + (0.357 s)*t  - 2.45 s2 = 0,
t = [-0.178 +(0.031 + 2.45)1/2]s = 1.4 s.
During this time the horizontal position of the mouse changes to x =  x0 + v0cosθ0t,
x = -4 m * (3.5 m/s)*0.866*(1.4 s) = 0.23 m = 23 cm.
The center of the nest is at x = 0 and the nest has a radius of 15 cm.  The mouse misses the nest.

### Simultaneous motion

Some problems involve the motion of two particles.  Often these problems require that the two particles meet at the same place at the same time.  An often stated problem is the monkey problem.

A monkey is hanging from a branch in a tree, a certain height h above the ground.  A zookeeper stands a distance d from the tree, with a banana in his hand.  The zookeeper knows that the monkey always lets go of the branch just as the banana is thrown to it.  In which direction must the zookeeper throw the banana, so that the monkey can catch it?

Analysis of the problem:

The position of a projectile as a function of time is given by r = v0t + ½gt2.  We can view the motion of the projectile as a superposition of two motions, a motion with constant velocity v0 in the initial direction and a downward motion with constant acceleration, like the motion of a freely falling particle.  If the projectile is aimed at the target and fired at t = 0, then motion with constant velocity v0 will bring the projectile to the initial position of the target at some later time t.  In the time interval between 0 and t the downward motion with constant acceleration carries the projectile downward by an amount ½gt2.  Superimposing the two motions will bring the projectile at time t to the position of the freely falling target at time t, independent of the magnitude of v0.  In the time interval between 0 and t the freely falling target moves downward by an amount ½gt2.

#### Demos:

Ball and cart have the same constant vx.  The ball's and the carts horizontal motions are the same.  The ball's vertical motion is independent of its horizontal motion and does not influence its horizontal motion.

Without gravity, the arrow would hit the original position of the target.  With gravity, arrow and target accelerate downward at the same rate.  (Youtube)