Projectile Help

[Greggler]

Hey guys just having a little trouble with some projectile questions at the moment.

Specifically its questions where something is launched off a cliff or raised surface of somekind, and they then ask you to state the total travel time of the object.

I know how to use the vertical velocity (making it = to 0) in order to find the maximum height, then doubling this value, but this only gives time for the parabola. The there is the extra time to compensate for the extra distance from where the ball is launched from.

In checkpoints it states to use the formula x = ut + 0.5gt2. And their value for u is is the vertical velocity multiplied by -1.
I can simply use this formula for the questions and get all of them right, but i would really like to know the theory behind this.

Sorry if my explanation is a bit dodgy, but any help would be greatly appreciated

[superflya]

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Vectors/ProjectilesMotion.html

that explains it pretty well, doesnt go into great detail so u wont fall asleep reading it oh n the formulas may look different to what u use but they are essentially the same thing.

[QuantumJG]

If it's only about finding the flight time (t_f), then the best formula to use is:

s = s₀ + ut + 0.5at²

now let's say we start up on some structure of height h above where it lands (let this be the zero point). The initial velocity of our projectile is u, the angle is θ and the acceleration is -g

0 = h + usin(θ)t - 0.5gt²

Now let's use the quadratic formula to find t and set out some restrictions by saying t_f > 0 and 0 θ 90^o  



Someone correct this if I'm wrong!
  

[Greggler]

thanks. i think that kinda makes sense.

its just annoying. i understand how to use this formula that checkpoints is giving me. (x = ut + 0.5gt^2)

but i wanna know how they make -u = the vertical velocity (sin(theta) x launch velocity)

[QuantumJG]

thanks. i think that kinda makes sense.

its just annoying. i understand how to use this formula that checkpoints is giving me. (x = ut + 0.5gt^2)

but i wanna know how they make -u = the vertical velocity (sin(theta) x launch velocity)

So you don't understand why the verticle component of the launch velocity is equal to the launch velocity times sin(θ)?

The reasoning is easy.

Consider the unit circle (I.e. x² + y² = 1), now it's easy to say that if we put some arbitrary point on the circle (x,y) so it's easy to see that x represents the horizontal distance from the origin and y represents the verticle distance from the origin. Now if we connect the dots from the origin (0,0) to (x,0) then (x,0) to (x,y) and finally (x,y) to (0,0) you get a right triangle where the adjacent = x, opposite = y and finally the hypotenuse is and it makes an angle θ with the x-axis.

Now,
 









I hope that helps explain why.