# I've forgotten the formula for distance thrown

• Tybras
In summary, the problem asks for the launch speed, the maximum horizontal distance and the maximum vertical distance an object thrown can cover, and how long it will 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. The attempt at a solution is to use the force, velocity, and position information to find the time of release and the ground distance.
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

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.

Tybras
Tybras said:
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?

RUber said:
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.

Tybras said:
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.

Tybras said:
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.)

## 1. What is the formula for distance thrown?

The formula for distance thrown is d = v2 / g, where d is the distance, v is the initial velocity, and g is the acceleration due to gravity.

## 2. How do I calculate the distance thrown?

To calculate the distance thrown, you need to know the initial velocity and the acceleration due to gravity, and then plug them into the formula d = v2 / g.

## 3. Can you give an example of how to use the formula for distance thrown?

Sure, for example, if you throw a ball with an initial velocity of 20 m/s and the acceleration due to gravity is 9.8 m/s2, the distance thrown would be calculated as d = (20 m/s)2 / (9.8 m/s2) = 40.8 meters.

## 4. Is the formula for distance thrown always accurate?

The formula for distance thrown is accurate in a vacuum, but in real-world scenarios, factors such as air resistance and friction can affect the distance thrown. Therefore, it may not always be completely accurate.

## 5. Can I use the formula for distance thrown in other situations besides throwing objects?

Yes, the formula for distance thrown can also be applied in situations such as calculating the distance a projectile travels when launched from a height or the distance a car travels when braking with a certain deceleration rate.

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