Please Help Me With This Physics Exercise

[mufarrid]

A chopper drops boxes of relief goods to a group of people stranded on an island. The goods were released 20.0 m above the sea and landed 40.0 m from a point exactly below where the goods were released. What was the velocity of the goods when they were released? What was the velocity of the chopper?

[Atheist]

I imagine that the velocity of the goods when released would be the same as the velocity of the chopper. I can't do the rest of the necessary calculation. Maybe someone else can help.

[bobbinwales]

Work out how long it takes to drop 20m due to gravity.
Then use that figure to work out how fast helicopter went at if it done 40m in the same time.

[bobbinwales]

That wont work actually because it travels in a diagonal si you may need pythogoras or signes and cosigner . Night school memories come flooding back

[Atheist]

bobb... after release I think it might follow a parabolic path (or some kind of curve, rather than a diagonal striaght line).

[bhg481]

I've just replied to this and my post has disappeared.

Bob's suggestion is OK. Once you've calculated the drop time you also get the time taken for the box to travel 40m horizontally (ignoring air resistance) and this allows you to calculate the forward velocity of the box. This, in turn, is equal to the forward velocity of the chopper.

[bhg481]

parabolic path is irrelevant. It's only parabolic because the box is accelerating, not dropping at a constant speed.

[Buenchico]

As Atheist say, the velocity of the goods at the point of release will be the same as that of the chopper. i.e. 19.81m/s parallel to the surface of the sea: https://ibb.co/Zxd7Z6j

[bobbinwales]

I vaguely remeber triangle of force's but yes maybe just need downwards speed and horizontal speeds sepaerate.
Good luck with the calculation

[Canary42]

I presume by "velocity" you mean the horizontal velocity. This will be equal to that of the helicopter. The vertical velocity will be nil at the point of release and will increase according to the acceleration due to gravity (32 foot per sec per sec when I was at school if I remember correctly, but you use whatever metric equivalent you have been taught. As has been stated, ignore air resistance for the object and the helicopter.) Then just follow bobbinwales' solution.

[Buenchico]

I want to know why the helicopter can't simply land on the island and pick everyone up ;-)

[bhg481]

Too many trees/soil too soft/would ruin test question.

[bobbinwales]

Covid restrictions on social distancing buenchico?

[Buenchico]

LOL @ Bob!

[Etch]

Looking at the values given, I suspect this is a case where the student is allowed to assume that g, the acceleration due to gravity, is 10 m/s² to make the calculation easier: t = 2 seconds and u = 20 m/s

[davebro]

"above the sea"

presumably the goods would land in the water (or at best on a beach!)

[Etch]

Perhaps the first question is supposed to be "what was the velocity of the goods when they landed": this would then require the student to show their knowledge of, and how to use, the equation
v² = u² + 2.a.s

[jourdain2]

This is where you need Jim.

[bhg481]

This is O level physics/applied maths - a bit elementary for Jim, I think.

[jourdain2]

Well, I accept that you would know, bhg, I certainly haven't a clue - but Jim is a teacher (lecturer I think) so he would be able to explain it.

[Buenchico]

^^^ I'm not even sure which Jim you're referring to. Both JimF and Jim360 have a scientific background.