Projectile Motion problem

djt0607

Hi,

I got this problem sheet from college and have been trying to figure it out for ages, but unfortunately I have been unsuccessful, so im hoping i can get some help here...ok heres the problem....(this problem does not contain any numbers, but the object is to purely produce general answers)

A projectile is launched from height h with initial velocity Vo at angle [alpha] with the horizontal. Find the law of motion and determine:

1. The max height reached by the projectile
2. The time to reach the max height
3. The horizontal distance at the max height
4. The total horizontal distance travelled
5. The time of flight

I know that the law of motion to use is Ro + Vot + 1/2at^2.
Ro = (0,h)
Vo = (Vox,Voy)
a = (0,-g)

I would be so so greatful if you could help me with any of this problem or if you could point me in the right direction of a decent website relating to projectile motion which uses the same convention as im using here.

Thanks.

Damo

LeixlipRed

Gentle nudges in the right direction are acceptable. Anything else and bye bye thread.

LeixlipRed

I think you need to review basic calculus. Derivatives, critical points, etc.

Yakuza

It's been 20 years since I did applied maths, but here goes:

Firstly I'm assuming you're familiar with the 3 equations of motion:
1) v = u + at
2) v² = u² + 2as
3) s = ut + ½at²

where:
u = inital velocity
v = final velocity
a = acceleration
t = time
s = distance covered

You need to split the motion into two components, the x (i) and y (j) planes. As you point out, there is no deceleration in the X direction, only negative g in the y direction

Imagine a right-angled triangle with angle α from the horizontal and a hypotenuse (long side) of Vo.

What proportion of Vo is going in the Y direction?
(Hint:
Swimmers Often Have
Cramps And Have
To Obtain Assistance )

Once you've worked out the Y component of the velocity (call it Vy), how long will it take the Y component of the velocity to reach zero under acceleration -g? This will give you the time to reach the max height (call it Ty), so you can now work out what the height is (call it Sy) - bear in mind that this is the height that it reaches from the gun, you have to add on h (the height of the gun from the ground).

If Ty is the time from when it leaves the gun to when it starts to fall, how long will the projectile be travelling for? (What comes up, must come down!) Call this Tx - don't forget that the projectile started from the gun at height h so it has to fall down a total of height Sy+h so Tx is not exactly double Ty.

Given that the horizontal component of the velocity is Vx, it is travelling for Tx seconds under zero acceleration, how far does it travel in the X plane (call this Sx).

I hope that helps, without spoonfeeding you the actual answers!