How much thrust is required to lift 300 pounds up to 100 feet?

Question

To clarify, let me ask this way, "How much thrust is required to lift 300 pounds up to 100 feet in 60 seconds, or rather how much thrust is required to hold 300 pounds at 100 feet?" I need to find out how much thrust would be needed to lift a 200 pound person wearing a 100 pound jet pack(weight will remain constant due to no fuel burn off for argument sake).

OK, If I understand correctly and put in simplest terms 301 Pounds of Thrust wil lift 300 pounds of Mass at an extremely slow rate, Increaseing thrust at that point will decrease the time to reach the given height. Is there a curve to be considered ie. as more height is achieved more thrust is required to maintain that height?

Answer 1

How fast to you want to get there? Faster acceleration requires more thrust for a given mass. So your question as given is unanswerable.

F, as in F = ma, is your thrust. f = F - W = ma - mg = m(a - g) = ma'; where f = the net force acting on the body, F is your thrust acting upward, and W is the downward acting weight of the thrusted body.

You can see if the thrust (F) is just equal to the body's weight (W), there would be no net acceleration (a'). The body would just sit there on the launch pad. On the other hand, if F >>> W, the mass would be accelerated at a very large a' and get to 100 feet in the blink of an eye.

One of the major considerations overlooked here is m. In the above equation, we assumed m to be constant...it is not in a rocket using fuel.

Fuel also has mass that has to be lifted by the thrust; so as it is burned to create the thrust, the rocket is actually losing mass...the mass of the fuel. The solution to this realistic thrust problem involves partial differential equations. I don't think you want to go there.

By the way, 30,000 foot-pounds is not the answer. That answer is work, not thrust, which is in pounds or Newtons. If you asked how much work does it take, 30,000 f-p would have been correct.

PS: Given the clarifications:

h = 1/2 a' t^2; where h = 100 ft, t = 60 sec

Use the above to solve for a' (which is the net acceleration after accounting for g) and plug that into f = F - W = ma'; using g = 32.2 ft/sec^2, m = W/g, and W = 300 lbs

Once you get to h = 100, to stay there you have to set a' = 0 and redo the equation F - W = 0 to solve for F (the thrust when net thrust f = 0).

You are wise to asume m is constant, it's messing otherwise.

Answer 2

Do the maths. A small plane only needs to flow about one hundred twenty mph. The go back and forth, on the top of the improve section needs to be going about 17000 mph. On good of it somewhat is the actual shown reality that the go back and forth weighs 2000 circumstances more beneficial than a Cessna Skymaster. And, of route, the go back and forth is doing a vertical takeoff, at the same time as a Cessna is doing a horizontal takeoff, making optimal use of its wing boost.

Answer 3

Anything over 300 pounds.

Answer 4

30,000 foot-pounds

Answer 5

u tell me.