a projecile problem (1 Viewer)

[lollol]

a missile was launched from the ground at an angle of 30, the initial velocity was 300 m/s

a)find the total flight time
b)what is the value of horizontal velocity at 20s

 

[denoz]

a missile was launched from the ground at an angle of 30, the initial velocity was 300 m/s

a)find the total flight time
b)what is the value of horizontal velocity at 20s

first u have to find the vertical and horizontal components:

vertical=300 sin30=150m/s
horivontal=300 cos30=260m/s

a) to get time to max hieght you use:
v=u+at
0=150+(-9.8)t note:v=0 as at max height there is no vertical velocity
t=15.3 seconds

therefore flight time = 2x time to max height=30.6 seconds

b) since horizontal velocity remains constant = 260m/s

hope this helps

 

[lollol]

thanks denoz, it's very helpful

 

[airie]

Or you could use the formula (delta)y = u_yt + 0.5a_yt² for (a), where (delta)y = 0, a_y = g = -9.8ms^-2 and u_y = u sin(theta) = 300(sin 30 degrees) = 150m/s, which works regardless of the point of launch

 

[Kabbasi]

Alternatively since you know the angle and the initial velocity, you could use the formula for time, i.e. time = 2u sin (angle) / gravity (9.8)

 

[twilight1412]

Alternatively since you know the angle and the initial velocity, you could use the formula for time, i.e. time = 2u sin (angle) / gravity (9.8)

ive never seen this formula before where did you get it?
u.sin(angle) = vertical velocity
vertical velocity/ gravitational acceleration = time ....
.... ahh right dw im an idiot ><
i didnt realise you used the time for the whole journey

nice derivation though i wouldnt have thought of it like that

 

[Forbidden.]

ive never seen this formula before where did you get it?
u.sin(angle) = vertical velocity
vertical velocity/ gravitational acceleration = time ....
.... ahh right dw im an idiot ><
i didnt realise you used the time for the whole journey

nice derivation though i wouldnt have thought of it like that

Don't worry, our teacher can derive LOTS of complicated formulas from other formulas they look like they belong to the Mathematics Extension 1 course.

Or you could use the formula (delta)y = u_yt + 0.5a_yt² for (a), where (delta)y = 0, a_y = g = -9.8ms^-2 and u_y = u sin(theta) = 300(sin 30 degrees) = 150m/s, which works regardless of the point of launch

Why is (delta)y=0 ? Does this have something to do with displacement ? I'm confused at what my teacher teaches me.

 

[twilight1412]

Don't worry, our teacher can derive LOTS of complicated formulas from other formulas they look like they belong to the Mathematics Extension 1 course.



Why is (delta)y=0 ? Does this have something to do with displacement ? I'm confused at what my teacher teaches me.

delta y in this case seems to be displacement
remember displacement and distance are 2 different things
i think thats the maths formula heres the physics one similar but clearer
s = (1/2)at^2 + ut

a = acceleration which in this case in gravitational
u = inital velocity
s = displacement of object
t = time
this can be derived from
v = u + at
and
2as = v^2 - u^2

 

[airie]

f3nr15, since the object was launched from the ground (and assuming that it landed on the ground ), the point of landing is level with the point of launch. Therefore, the displacement in the vertical direction, ie. (delta)y, is zero

And no twilight1412, that formula I mentioned is actually the "physics one" with the "physics notations" for the variables, as in the physics syllabus

 

[Forbidden.]

f3nr15, since the object was launched from the ground (and assuming that it landed on the ground ), the point of landing is level with the point of launch. Therefore, the displacement in the vertical direction, ie. (delta)y, is zero

And no twilight1412, that formula I mentioned is actually the "physics one" with the "physics notations" for the variables, as in the physics syllabus

k thnx in all cases d_y is always 0.

delta y in this case seems to be displacement
remember displacement and distance are 2 different things
i think thats the maths formula heres the physics one similar but clearer
s = (1/2)at^2 + ut

a = acceleration which in this case in gravitational
u = inital velocity
s = displacement of object
t = time
this can be derived from
v = u + at
and
2as = v^2 - u^2

Since you use s as your displacement instead of the d we use, here are the five equations of motion I remember:

1. s/t=v+u/2
2. v=u+at
3. 2sa=v²-u²
4. s=ut+1/2at²
5. s=vt-1/2at²

As for two-dimensional projectile motion, I assume the y-displacement formula to be:

d_y=u_yt+1/2a_yt²

And assume the x-displacement to be:

d_x=u_xt+1/2a_xt²


And when finding out the displacement between the bar and athelete in high jump at the instantaneous moment the athelete is on top, for example:

d_y=v_yt-1/2a_yt²

Where v_y = 0 because at the vertex, velocity is 0.

 

[airie]

k thnx in all cases d_y is always 0.

Unless, of course, the object ends up lower than the starting point