projectile (1 Viewer)

[synthesisFR]

can someone do part c and onwards my tutor did a wierd method i dont understand i did it differentyy

 

[synthesisFR]

helo

 

[scaryshark09]

 

[scaryshark09]

this is what i would have done. I am not confident in my answer to (c) (i just tried something random cause i had no idea). I think the others are right tho (let me know if they aren't)
could you let me know what the answer is to (c)?

 

[synthesisFR]

this is what i would have done. I am not confident in my answer to (c) (i just tried something random cause i had no idea). I think the others are right tho (let me know if they aren't)
could you let me know what the answer is to (c)?

its like 4.6

 

[scaryshark09]

yeh ok, sorry idk how to do (c).
are my other answers right?

 

[synthesisFR]

yeh ok, sorry idk how to do (c).
are my other answers right?

yep

 

[carrotsss]

yep

The overall velocity in the horizontal plane is sqrt(30^2+40^2)=50

So the angle is arctan(4/50)=4.6 degrees below the horizontal

 

[synthesisFR]

The overall velocity in the horizontal plane is sqrt(30^2+40^2)=50

So the angle is arctan(4/50)=4.6 degrees below the horizontal

wait but i thought it landed on level with the control tower?

 

[scaryshark09]

wait but i thought it landed on level with the control tower?

Yeh it does. When you find the velocity 50 part that it on ground level and the 4 velocity is the vertical part.

 

[scaryshark09]

how come the velocity is like constant tho? Wouldn’t the plane be decelerating?

 

[synthesisFR]

Yeh it does. When you find the velocity 50 part that it on ground level and the 4 velocity is the vertical part.

wait im having an aneurysm ik this is a direction of motion question but wtf why is there a vertical part if its on the grand the the tower is on the ground huhhh

 

[scaryshark09]

wait im having an aneurysm ik this is a direction of motion question but wtf why is there a vertical part if its on the grand the the tower is on the ground huhhh

When you find the angle you have to use velocity in y direction and velocity in x direction and use the formula tan(theta) = y/x where y and x are velocities. There’s no way you can find angle without the vertical part. Like that’s essentially the velocity in the vertical direction as soon as it touches the group at that exact moment. Soon after it will be 0.

 

[synthesisFR]

When you find the angle you have to use velocity in y direction and velocity in x direction and use the formula tan(theta) = y/x where y and x are velocities. There’s no way you can find angle without the vertical part. Like that’s essentially the velocity in the vertical direction as soon as it touches the group at that exact moment. Soon after it will be 0.

yeah but this is 3d projectile so where is the z compoentn if its landed i thought u just find it from the horzontal x axis or smth in that case

im gonna just sleep and redo this tmrw bc my brain cannot process this rn for some reason

 

[carrotsss]

wait but i thought it landed on level with the control tower?

the control tower is on the same level as the runway - and it wouldn’t matter if it wasn’t anyway, the velocity is constant, and so the velocity at the instant it reaches the runway before it gets levelled out is just the normal constant velocity, and hence the angle of the plane compared to the runway is also that. here’s a diagram that is questionably helpful

 

[synthesisFR]

the control tower is on the same level as the runway - and it wouldn’t matter if it wasn’t anyway, the velocity is constant, and so the velocity at the instant it reaches the runway before it gets levelled out is just the normal constant velocity, and hence the angle of the plane compared to the runway is also that. here’s a diagram that is questionably helpful
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im so dumb i was imagining the plane landing horizontally