I've forgotten the formula for distance thrown

[Tybras]

Homework Statement

I am to figure out the maximum vertical distance and the maximum horizontal distance for an object thrown, as well as how long it take to get there. The information given is the weight of the object, the pounds of force used to launch the object, the distance for the total acceleration for the launch, and the time it takes for the initial launch. I am told to assume that the wind will have no effect on the object for the duration of the flight. I have to use this on multiple problems and have been out of school for so long i have forgotten the equations.

first example is a 20 lb object is launched with a initial force of 1000 lbs that takes 7.5 seconds to go from 0 lbs of force to 1000 lbs of force across a distance of 30.66 inches. it is released from 88.33 inches off the ground.

I am to find how high it goes if it is thrown straight up, and how long it takes to hit the ground.

I am also to find, if thrown at the best possible angle, what is the maximum horizontal distance it could cover, how long it takes to cover that distance, and what is the maximum height it reaches at that angle.

Homework Equations

i have forgotten the formulas

The Attempt at a Solution

please help

 

[RUber]

Hello Tybras,

Force = mass x acceleration.
Acceleration is change in velocity / change in time.
Velocity is change in position / change in time.

Can you find the velocity at the time of launch?
Once you have that, the formula for the ballistic path is based entirely on initial velocity and position, since once it is released from the launcher, only gravity is acting on the object.

Acceleration of gravity, g = -32ft/sec or so.
Velocity, v= $\int_{t_0}^t g d\xi = g(t-t_0)+v_0$.
Assume $t_0 = 0$
Position, r = $\int_{t_0}^t v(\xi) d\xi = \int_{0}^t g\xi +v_0 d\xi = \frac{g}{2}t^2 +v_0 t + r_0$

So, if you know $v_0$ and $r_0$, your initial velocity and position, you only need to solve the quadratic for t to find when r = 0, i.e. back on the ground.

 

[haruspex]

force of 1000 lbs that takes 7.5 seconds to go from 0 lbs of force to 1000 lbs of force across a distance of 30.66 inches

This seems to be saying the force increases during the launch, but doesn't give us a way to know what the force is at all points. That makes it impossible to determine the launch speed. Is this the exact wording?

Acceleration is change in velocity / change in time.

That only works for constant acceleration, which does not seem not be what we have in the launch phase.

 

[Tybras]

from the way the problem is worded it seems that the acceleration is constant, reaching the final velocity that the 1000lbs of force would produce after traveling the distance of 30.66 inches in 7.5 seconds, so the initial velocity would be 0, with the final velocity of the launch having to be calculate, and used as the initial velocity value for the launch.

 

[SammyS]

from the way the problem is worded it seems that the acceleration is constant, reaching the final velocity that the 1000lbs of force would produce after traveling the distance of 30.66 inches in 7.5 seconds, so the initial velocity would be 0, with the final velocity of the launch having to be calculate, and used as the initial velocity value for the launch.

If the force is changing, then so is the acceleration. Therefore, acceleration is not constant.

 

[haruspex]

from the way the problem is worded it seems that the acceleration is constant, reaching the final velocity that the 1000lbs of force would produce after traveling the distance of 30.66 inches in 7.5 seconds, so the initial velocity would be 0, with the final velocity of the launch having to be calculate, and used as the initial velocity value for the launch.

Please provide the exact wording of the problem. (I'm wondering if the bit about the force starting at zero is something you have added.)