Theoretical description about free fall and acceleration

[atxjoe512]

Hey everyone... I have a question about free fall and acceleration.


I recently had a question on an exam that went something like this. "A ball is thrown directly up from a building 59.4 meters high. It barely misses the building on its way down and hits the ground 4 seconds after being thrown. What is the final velocity of the ball?"

I got the answer correct through a plug and chug method of using the formulas. However, I don't understand.

How does one simple formula account for both the ball moving up and then coming back down? Why don't you have to use two formulas... The first showing the final height after the ball was thrown from a height of 59.4m, and the next formula taking that maximum height to calculate the balls final velocity when it hits the ground.

I'm confused how the formula incorporates both the going up and going down of the ball.

Please help! I'd like a conceptual description if possible.

 

[cpscdave]

The kinematic equations provide a description of the behavior of an object when a constant force is applied, in this case the force of gravity.
From this the velocity of the object and as an extension the position of the object can be determined.

Why does it work for both "up" and "down". Well to confuse the issue the ball is only ever traveling in one direction. When the ball is thrown it is traveling "up" with a positive velocity. Just before it hits the ground the ball is still traveling "up" just this time it has a negative velocity.

 

[ehild]

You get the relation between height of the ball and the time of flight by assuming constant acceleration. It is just a second order function of the time, like a parabola. You can describe both branches - rising and descending- of a parabola, with the same formula y=ax²+bx+c, don't you? In case of a vertically thrown ball, y =g/2 t² + v₀t+ h, (v₀ being the initial velocity and h is the initial height). Plotted on the t-y plane, it is a parabola.

ehild

 

[Adjoint]

How does one simple formula account for both the ball moving up and then coming back down?

In that specific example you have a 1D vertical motion and you are wondering why same formula works for both moving up and down.
A similar problem can be a car moving forward and then backward.
In these problems it doesn't matter whether a particle is moving forward - backward or up - down, what matters is the total displacement.
Once you set up your coordinate system and defined positive and negative directions, you just substitute the initial position from the final position of the particle to get the displacement. And in all those 1D kinematic equations (like y = v₀t + 0.5t²) you just use this displacement (here y). Its not important what the particle is doing between its initial and final positions.
Going up or going down, a formula that uses displacement is enough.

 

[HalfLostAndSearching]

I can provide a theoretical explanation for free fall and acceleration to help clarify your confusion. Free fall is the motion of an object under the influence of gravity alone, without any external forces acting on it. In this scenario, the ball is thrown up from a building and is in a state of free fall as it falls back down towards the ground.

Acceleration, on the other hand, is the rate of change of an object's velocity over time. In the case of free fall, the acceleration is due to the force of gravity pulling the object towards the ground. This acceleration is constant and is equal to 9.8 meters per second squared on Earth.

Now, let's break down the scenario given in the question. The ball is thrown up from a height of 59.4 meters and reaches a maximum height before falling back down. During this time, the ball is experiencing a positive acceleration as it moves upward due to the initial force of the throw. However, as soon as the ball reaches its maximum height, it starts to fall back down towards the ground and experiences a negative acceleration due to the force of gravity. This results in the ball's velocity decreasing until it reaches the ground.

The formula used to calculate the final velocity of the ball takes into account these two phases of motion - going up and coming back down. It is a combination of the equations for displacement (d = vt + 1/2at^2) and velocity (v = u + at), where d is the displacement, v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

In the first phase, the initial velocity (u) is the velocity at which the ball is thrown, and the acceleration (a) is the constant acceleration due to gravity. This gives us the maximum height reached by the ball. In the second phase, the initial velocity (u) is the velocity at the maximum height (which is 0 m/s), and the acceleration (a) is still the constant acceleration due to gravity. This gives us the final velocity of the ball when it hits the ground.

So, to answer your question, the formula takes into account both the going up and coming back down of the ball by considering the different phases of motion and the corresponding values of initial velocity and acceleration.

I hope this explanation helps to clarify your confusion. Feel free to ask any further questions if needed. As a scientist, it is