projectile qs (1 Viewer)

[Danger]

eg. a projectile was launched at blah blah from a tower 500m above the ground, find the max. height reached by the projectile.

upon finding the max. height, do i also need to add 500m to the answer? thanks

 

[helper]

It depends on how the question is worded. If unsure make a statement at the end like:

---- m above tower or ---- + 500m above ground

 

[Forbidden.]

Path of trajectory ?

 

[Danger]

helper are we allowed to put that as an answer in the exam?

forbidden, yes


but lets say, the projectile is a firecracker and it explodes at a height of 300m above the ground. would this make any difference to the answer?

also, another question: why is it necessary to import a sideways thrust to a satellite?

I couldnt find anything on that q in books

 

[helper]

Yes because all you are doing is adding extra correcct money.

Remember the equations are always measuring from your starting position. Not sure what you mean by your firecracker question.

The satellite needs to receive a sideways thrust to turn it to its tangential circular velocity.

 

[Danger]

thanks.

when asked for 'the speed of the projectile at its maximum height', its just v=0, right?

 

[veloc1ty]

when asked for 'the speed of the projectile at its maximum height', its just v=0, right?

No, vy = 0 but there's still a horizontal component.

 

[Danger]

how would you find that horizontal component?

 

[Mark576]

Range = horizontal velocity * time;

therefore: horizontal velocity = Range / time

Note that horizontal velocity remains constant throughout an entire trajectory.

Range is the horizontal distance travelled. Alternatively you could use:

horizontal velocity = u cos @, where u is the total initial velocity and @ represents the angle of inclination to the horizontal.