A rock is released from rest on a 700 m building and time is 11.95 what is the speed and velocity of the ball just before it hits the ground? I know you have to use the Vf=Vo+at formula but I was wondering why can I not use this formula as well: Vf=Vot+1/2at^2. What's the difference
Oh my !!!
Vf=Vot+1/2at^2. NO !!!!
That is incorrect !
That is DISTANCE not speed
a = - g
v = Vo + a t like that
x = Xo + Vo t + (1/2) a t^2
I sure hope you check this site this morning !
Sorry I did not see this last night. It is important that you understand this. Please check it.
Damon
Thank you!
Oh, I see you’re trying to rock the physics world with your questions! Let's break it down with some Clown Bot humor!
The formula Vf = Vo + at is used to calculate the final speed of an object when it is given an initial velocity (Vo), acceleration (a), and time (t) traveled. It assumes that the object starts with an initial velocity and maintains a constant rate of acceleration.
On the other hand, the formula Vf = Vo + 1/2at^2 is used to calculate the final velocity of an object when you know the initial velocity (Vo), acceleration (a), and the time (t) traveled. However, this formula assumes that the object starts from rest (Vo = 0) and uniformly accelerates over the given time.
In your scenario, since the rock is released from rest, it's appropriate to use the simpler formula Vf = Vo + at. This formula works perfectly fine for determining the final velocity of the rock just before it hits the ground.
As for velocity, it is a vector quantity that includes both speed and direction. So, when the rock is falling straight down (assuming positive is downward), the velocity will have a magnitude equal to the speed but a negative sign to indicate the downward direction.
So, to sum it up, for your problem, you can use Vf = Vo + at to find the final speed, and the velocity just before it hits the ground will be the final speed with a negative sign to indicate the downward direction. Hope this answer rocks your world!
Both formulas, Vf = Vo + at and Vf = Vo + 1/2at^2, are derived from the kinematic equations of motion. The difference lies in the specific scenarios for which each formula is applicable.
Vf = Vo + at is used when the initial velocity (Vo) and acceleration (a) are constant throughout the motion. This equation assumes that the object starts from rest (Vo = 0). It is generally valid for situations where acceleration is constant, such as the motion of an object falling freely under gravity above the surface of the Earth.
On the other hand, Vf = Vo + 1/2at^2 is derived from a more general form of the kinematic equations called the second equation of motion, which applies to cases where initial velocity and acceleration are constant but time (t) can vary. This equation is commonly used for objects undergoing uniformly accelerated motion, such as a projectile launched from a horizontal surface.
In the given scenario of a rock falling from a building, it is appropriate to use Vf = Vo + at formula, as the acceleration due to gravity remains relatively constant near the Earth's surface. The value of acceleration can be taken as -9.8 m/s^2 since the rock is falling downwards. By substituting the given values, we can calculate the final velocity (Vf) of the rock just before it hits the ground.
To calculate the velocity using Vf = Vo + at, you need to know the initial velocity (Vo) and the time (t). The time can be obtained from the information given, which is 11.95 seconds. However, the initial velocity is not provided in the question, so further information is needed to solve for the exact velocity.
The velocity can be calculated by using the kinematic equation Vf^2 = Vo^2 + 2ad, where d represents the distance traveled (700 m in this case) and a is the acceleration due to gravity (-9.8 m/s^2). By rearranging the equation, you can solve for Vo:
Vo = √(Vf^2 - 2ad)
Once you have the initial velocity, you can use Vf = Vo + at to find the final velocity.
Remember to consider the negative sign for the velocity value, as the rock is falling downward towards the ground. Velocity is a vector quantity, and in this case, it will have a negative value to indicate the downward direction.