Help!! (1 Viewer)

[ghost-blade]

when using this formula for projectile motion, what do i use as theta, do i add the angle given to the 20?
in an example were the angle is 15 n u=30 what would i tpye in the calculator to solve for the range (d)?

 

[adnan91]

when using this formula for projectile motion, what do i use as theta, do i add the angle given to the 20?
in an example were the angle is 15 n u=30 what would i tpye in the calculator to solve for the range (d)?

Range = int horizontal velocity x time of flight

ux = vcos @ Personally i wouldnt use that formula. Not sure if your allowed to but i dunoo

 

[ghost-blade]

I don't think you absolutely need to use this formula, much less know it. There are other ways of finding the answer, even if they take more steps.

In this case, you double the angle of projection, 15 degrees, and substitute it into the formula.

thanx, so u always double the given theta to use in this formula?
so if its 30 il put (Ux Sin 60)/9.8 ? (just making sure)

reason i asked is cause there's no other way i can see to calculate the initial velocity if they only give you the range and angle of projection.

oh and i think its fine to use in HSC as in Q3 from the space topic in Success one Physics they use this formula to solve the Question.

 

[I-Love-Jesus]

You will never need to use that formula. Having said that, it can be a nice shortcut if projectiles aren't your strong point. One negative thing with it is that if it is projected from a different height to where it lands, that formula doesn't work. So I would suggest instead knowing how to derive it.

All you have to do is find the time of flight (by making y=0 or whatever the final height of the projectile is, and then solving for t), and then substitute that into R=u_xt.

In the case of the height of the projectile being the same at the beginning and end of its flight, you do this by:
y = u_yt -0.5*g*t²
0 = u_yt -0.5*g*t²
0 = t(u_y-0.5*g*t)
Therefore t=0, or t=2*u_y/g
since t=0 is the time of projection, the time of flight is given by t=2u_y/g
Substituting this into R=u_xt gives:
R=u_x(2u_y/g)
R=2u_xu_y/g
Since u_x=Vcos@, and u_y=Vsin@,
R=2(Vcos@)(Vsin@)/g
R=V²*2cos@sin@/g
Since sin(2@) = 2sin@cos@,
R=V²sin(2@)/g, which is your equation.

In the equation:
V is the initial velocity of the projectile (not purely horizontal or vertical, but a combination of both)
@ is the angle of projection, e.g. if it is fired at 15^o, you would enter @ as 15^o, and sin(2@) = sin(30^o)
g is the acceleration due to gravity, which on in this course is given as 9.8m.s^-1 on Earth.

 

[obimoshman1234]

when using this formula for projectile motion, what do i use as theta, do i add the angle given to the 20?
in an example were the angle is 15 n u=30 what would i tpye in the calculator to solve for the range (d)?

what 20??????

all i see is 2 times theta??????????

 

[Teh Duke]

In an exam, are we allowed to use that formula straight away or do we have to derive it.

 

[obimoshman1234]

well i dont know dont think u have to derive but it can only be used when projectile lands level with starting point.


plus i wouldnt use it i just prefer using R=ux t lol i hate notationing here