Solving Projectile Motion: Finding Max Height & Final Speed

In summary, For a force F = −mg ± cv^2 (the sign is chosen so the drag force is opposite the motion), you can use a trick: let u = v^2, and generate a differential equation for u(x) (not u(t)!), which is not too hard to solve.
  • #1
Gogsey
160
0
For a force F = −mg ± cv^2 (the sign is chosen so the drag force is opposite the motion), you can use a trick: let u = v^2, and generate a differential equation for u(x) (not u(t)!), which is not too hard to solve. (Hint: the chain rule gives this dv/dt=dv/dx * dx/dt should help.)
a) Solve this to show that a projectile thrown vertically upwards from the ground with speed v0 reaches a maximum height y=V(t)^2/2g*(ln(1 + V(o)^2/V(t)^2, where V(t) is the terminal speed.
b) Show that the projectile hits the ground with speed V(f) given by:
V(f)^-2=V(o)^-2 + V(t)^-2

So, part a) is ok, I have that after quite a bit of math, but I'm a little lost on b). When a projectile hits the ground, that's usually at t((y) = 0, but there is not time term in this equation.
 
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  • #2
So let's call T the time when the projectile hits the ground.
What do you know about the value of y(T)?
The final velocity is V(T) = V(f).
This should give you an equation.

Also, please check the answer you are supposed to derive... is there really a V(t) in there?
 
  • #3
CompuChip said:
Also, please check the answer you are supposed to derive... is there really a V(t) in there?

I obtained the desired result, so I suspect that you may have used the wrong sign in front of 'c' during the falling portion of the motion.
 
  • #4
Gogsey said:
So, part a) is ok, I have that after quite a bit of math, but I'm a little lost on b). When a projectile hits the ground, that's usually at t((y) = 0, but there is not time term in this equation.

I hope you aren't going to all the trouble of actually finding x(t) (or y(t) if that's what you're calling it)!

All you really need to do is find [itex]u(x)=v^2(x)[/itex] as the projectile falls from a height of [tex]x=\frac{v_t^2}{2g}\ln\left(1+\frac{v_0^2}{v_t^2}\right)[/tex] to [itex]x=0[/itex]. The impact speed will simply be [itex]v_f=v(x=0)=\sqrt{u(x=0)}[/itex]
 
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Related to Solving Projectile Motion: Finding Max Height & Final Speed

1. What is projectile motion?

Projectile motion is the motion of an object through the air, influenced only by the force of gravity. It follows a curved path known as a parabola.

2. How do you find the maximum height of a projectile?

To find the maximum height of a projectile, you can use the equation h = (v2sin2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

3. What is the final speed of a projectile?

The final speed of a projectile is the speed at which it hits the ground. It can be calculated using the equation v = √(v02 + 2gh), where v0 is the initial velocity, g is the acceleration due to gravity, and h is the height at which the projectile lands.

4. How do you solve for projectile motion with air resistance?

To solve for projectile motion with air resistance, you would need to use more complex equations that take into account the effects of air resistance. These equations involve factors such as the density of the air, cross-sectional area of the object, and drag coefficient.

5. What are some real-world applications of projectile motion?

Projectile motion is used in various real-world applications, such as sports like basketball and baseball, where the trajectory of the ball follows a parabolic path. It is also used in engineering and physics, such as in the design of rockets and satellites. Additionally, it is used in military operations, such as determining the trajectory of missiles and projectiles.

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