i just want to know form where i got on this formula the 4.9t² on this formula s=4.9t²+Vot+h because the book only say that a physics theory shows that when an object is thrown upward with an initial velocity Vo, its approximate height is given by that quadratic function
i will appreciate if you can explain me that because i have a math problem that uses that equation but i don't know from wheres that 4.9t²
It comes from calculus, and I assume you do not know that.
Newton found the velocity of a falling object was proportional to the time it fell.
vincreased= constant*time
but the constant we now call g, 9.8m/s^2, which comes from his universal gravitational equation.
From this, we get
Vfinal=vinitial+ g*time
Now the distance something falls is the average velocity, times time.
distance=(vfinal+vinitial)/2 * time
d=(1/2 (vinital+ g*time + vinitial)*time
= 1/2 (2vinitial*time + g*time^2)
= vinitial*time + 1/2 g*time^2
of course, 1/2 g = 4.9 m/s^2
The value 4.9t^2 comes from the acceleration due to gravity. In physics, when an object is thrown upwards or falls down under the influence of gravity, it experiences a constant acceleration. The standard value for acceleration due to gravity is approximately 9.8 m/s^2. However, in this quadratic function, it is divided by 2 to account for the fact that the object is only traveling in one direction (upwards or downwards).
To understand where the 4.9t^2 comes from, let's break down the equation step-by-step:
1. The general equation for displacement (height) with respect to time is given by s = Vot + (1/2)at^2 + h.
2. In the equation, s represents the displacement or height, Vo is the initial velocity, a is the constant acceleration, t is time, and h represents the initial height.
3. Since the object is thrown upwards under the influence of gravity, the acceleration a is considered negative, because gravity is acting against the motion. Therefore, a = -9.8 m/s^2.
4. Substituting the values into the equation, we get s = Vot - (1/2)(9.8)t^2 + h.
5. Simplifying further, we find that s = Vot - 4.9t^2 + h.
6. The 4.9t^2 term is the result of multiplying (1/2)(9.8)t^2, or (1/2)at^2, which accounts for the effect of gravity on the object's height.
Therefore, the 4.9t^2 term in the equation s = 4.9t^2 + Vot + h arises from the acceleration due to gravity, which is modeled as -9.8 m/s^2 in the upward direction.
The term 4.9t² in the equation s = 4.9t² + Vot + h comes from the physics concept of gravity and the motion of objects in freefall.
When an object is thrown upward with an initial velocity Vo, it eventually reaches its highest point and then starts to fall back down due to the force of gravity. During the time the object is in the air, gravity continues to act on it, causing it to accelerate downward.
The acceleration due to gravity on Earth is approximately 9.8 m/s². However, since we are looking at the height of the object, we only consider the vertical component of the object's motion. Due to this, the acceleration term in the equation is divided by 2.
Hence, the 4.9t² term represents the change in height of the object at time t due to gravity. It is derived from the formula for the displacement of an object in freefall, S = V₀t + ½at², where V₀ is the initial velocity, t is the time, a is the acceleration, and S is the displacement or height.
By considering only the vertical component of motion, we use 4.9 instead of 9.8 as the acceleration term, which is why the equation becomes s = 4.9t² + Vot + h.
So, the 4.9t² term represents the height the object gains or loses due to gravity during the time t.