Thread: Dynamic Help

I'm just having trouble with a few problems and really would appreciate some help:

Projectile Motion
1. A bullet is fired with a velocity whose vertical and horizontal components are Vy and Vx respectively. If the horizontal component of velocity is 600 m/s, find the angle of projection that must be used for the bullet to hit a target 2m above the muzzle and 500m distant from it.

Simple Harmonic Motion
2. A particle starts from rest at x = 10 and proceeds with simple harmonic motion about x = 0 so that after 2 seconds it reaches x = 5. Find:
(a) an expression for the displacement (x) a any time t.
(b) the speed at x = 0
(c) the amplitude and period of motion
(d) the maximum speed
(e) the maximum acceleration

Thanks!

Question 2.

Since you are talking about simple harmonic motion the equation of motion will have the form:

either:


or



You can take your pick between the two, but I will assume the first option.

1. If you substitute X=0 and y=10 into the equation then you can solve for A.
2. Velocity can be found by differentiating with respect to t. Now you can substitute t=0 and y'(0)=0 and solve for B (velocity at time = 0 is zero).
3. Substituting y=5 and the time at which this occurs will then allow you to solve for w.
4. At this point you have a complete equation of motion and can find the position at any time t.
5. Differentiating your equation for motion will give you the velocity at any time t.

Originally Posted by holmesb

Projectile Motion
1. A bullet is fired with a velocity whose vertical and horizontal components are Vy and Vx respectively. If the horizontal component of velocity is 600 m/s, find the angle of projection that must be used for the bullet to hit a target 2m above the muzzle and 500m distant from it.

The key is that the horizontal component of motion is given, the horizontal distance is 500m and in the ideal case the horizontal component of velocity will not change. Therefore the flight time will be 500/600.

Then by linearity:



and

Oh bother. They gave the horizontal component of the velocity. Arg!

Thanks for the catch Kiwi_Dave.

-Dan