Projectile Motion help? (1 Viewer)

[aerialsprite]

Could anyone please explain to be how to find the angle and velocity of impact?

 

[shafqat]

Could anyone please explain to be how to find the angle and velocity of impact?

mmm i think you get time of impact by putting y = 0
Then put that time into the equations for vertical and horizontal velocity to get those
Then draw a vector triangle to find the combined velocity and the angle of impact

 

[m_isk]

If the projectile motion is a purely even parabola (ie if starts at the ground and finishes at the ground) then the magnitude of the velocity of impact will equal the maginitude of the vlelocity of launch. For the direction, it will just be the opposite as launch (ie instead of x degrees ABOVE the horizontal, it will be x degrees BELOW).

 

[aerialsprite]

If the projectile motion is a purely even parabola (ie if starts at the ground and finishes at the ground) then the magnitude of the velocity of impact will equal the maginitude of the vlelocity of launch. For the direction, it will just be the opposite as launch (ie instead of x degrees ABOVE the horizontal, it will be x degrees BELOW).

What if you're projecting something h metres above the ground?

 

[nick1048]

for all verticle velocities i.e. u = Vsin alpha
you use the formulas of physics i.e.
v = u + at
v^2 = u^2 + 2as
s = ut + 0.5at^2

and if your trying to find the total range it's handy to know
d = s x t
i.e.
horizontal distance = horizontal velocity (Vcos alpha) multiplied by time of flight

The formulas are rendered useless when you don't have a perfect parabola and you have to use these methods and the parametric ones to solve for the angle.
Goodluck hope it helps.

 

[shafqat]

you cannot use any of the formulae from physics, you will not get the full amount of marks if you do and get the correct answer
the only equations you can assume are horizontal acc = 0, vertical acc = -g, unless the question says you can assume the other equations of motions (velocity, displacement)
as i posted above, find the time by putting y = 0
so gt^2/2 = vtsin@
t = 2vsin@/g
substitute that into the equations for vertical and horizontal velocity, and then draw a vector triangle, and find the velocity using pythagoras' theorem.
this will work even if you throw from a height. in that case you have to use the quadratic formula to find t when you put y = 0

 

[m_isk]

What if you're projecting something h metres above the ground?

then it won't work. plus if you just write this you won't get full marks in any exam. its just a handy rule of thumb to see if you've done silly mistake.

 

[nick1048]

vector triangle is another physics method... and I don't know how to do that so that's really useless.

 

[aaaman]

Could anyone please explain to be how to find the angle and velocity of impact?

the formulas of

Ux = s cos#

Uy = s cos#

# - theater, i cbf getting da real symbol

 

[acmilan]

The way I do it:

You use for formal for horizontal and vertical velocity, and sub in the value for time when the object impacts. This will give the horizontal and vertical velocity at the instant it impacts. Then construct a right angle triangle, the vertical side is vertical velocity, the horizontal side is horizontal velocity, the hypotenuse will be thevelocity on impact and then you can work out the angle needed by basic trig ratios