A series of separate differential equations can be set up using the diagram shown below, based on projectile motion - incorporating gravity but neglecting air drag effects.

**Diagram depicting projectile motion in 2 dimensions.**

v_{ x} = dx/dt = _{o} cos(q)

v_{y} = dy/dt = _{o} sin(q)

The force components at time t are:

F_{ x} = 0 and: _{ y} = - mg

With these incorporated the differential equations become:

i) m d^{2}x/dt ^{2 }= 0 and ii) m d^{2}y/ dt ^{2 }= - mg

Each equation above requires two integrations, yielding 4 constant of integration in all.

For eqn. (i) we get on integrating: dx/ dt = c1 Then:

ò dx = c1 òdt

Yields: x = c1t + c2

Here for eqn. (ii) we have:

d^{2}y/ dt ^{2 }= - g So: dy/dt = - gt + c3

And: y = - ½ g t^{ 2} + c3t + c4

Based on the initial conditions given we have:

c1 = v_{o} cos(q), c2 = 0, c3 = _{o} sin(q), c4 = 0

Then the position of the projectile at time t seconds after firing can be found from:

x = v_{o} cos(q) t and: y = - ½ g t^{ 2} + v_{o} sin(q) t

Clearly, the projectile attains maximum altitude when its y -component of velocity is 0, i.e.

dy/ dt = 0 = -gt + v_{o} sin(q)

This must occur at time:

t' = v_{o} sin(q)/ g

Then the maximum altitude is:

y _{max} = - ½ g t'^{ 2} + v_{o} sin(q) t' =

- ½ g [v_{o} sin(q)/ g]^{ 2 }+ v_{o} sin(q) [v_{o} sin(q)/ g]

= v_{o} sin(q) ^{2 }/ 2 g

Example Problem:

Find:

i) The maximum height attained when the launch angle for a projectile is q = 30 degrees.

ii) Write an equation for the slope (dy/dx) of the path of the projectile at any point.

iii) Write the condition to obtain the range R (maximum horizontal distance) of the projectile and find it.

iv) From (iv) what can be inferred about the angle F ?

__Solutions:__

__i)__ The maximum altitude is:

y _{max} = - ½ g t'^{ 2} + v_{o} sin(q) t' =

- ½ g [v_{o} sin(q)/ g]^{ 2 }+ v_{o} sin(q) [v_{o} sin(q)/ g]

= v_{o} sin(q) ^{2 }/ 2 g

q = 30 degrees:

= v_{o} sin(30) ^{2 }/ 2 g

But sin (30) = 1/2 so:

y _{max} = v_{o} (1/2) ^{2 }/ 2 g

= v_{o} (1/4) / 2 g = v_{o} / 8 g

ii) Slope at any point:

dy/dx = (dy/dt)/ (dx/dt) =

- ½ g t^{ 2} + v_{o} sin(q)/v_{o} cos(q)

iii) Range (R) occurs when y = 0

This occurs at *double the time* for max. altitude since trajectory is symmetric.

So: 2t' = 2 ( v_{o} sin(q)^{ }/ g )

Then x = R = v_{o} cos(q) 2 ( v_{o} sin(q)^{ }/ g )

= ( v_{o} ^{2 }/ g) sin(2q)

iv) The slope dy/dx is (expanded from (ii)):

dy/dx = -2 v_{o} sin(q) + v_{o} sin(q)/ v_{o} cos(q)

= -_{ }[sin(q)/ cos(q)] = - tan (q)

This shows the angle F = p - q

i.e. projectile returns to Earth at same angle it left.

** Suggested Problem**:

^{o}, k = 0.01mv v = 10

^{5}fs

^{-1}and t = 10s

## No comments:

Post a Comment