In kinematics, there are 4 MOST COMMON KINEMATIC FORMULAS we use to solve the problems.

These are

1. d = (Vi)(t) + (1/2)(a)(t^2)

2. Vf = Vi + at

3. (Vf)^2 = (Vi)^2 + 2ad

4. d = ((Vi +Vf)/2)(t)

But when do we really use them in solving the problems?

Let us first identify the variables involved in the equations.

The first variable "d" is the distance that the certain object in motion reaches after a certain amount of time (t).

Vf and Vi are what we call the FINAL AND INITIAL VELOCITIES of the object. Why there is final and initial. That is, in kinematics, we are to observe the objects in "motion". So some problems may say that the object has moved from a certain velocity and stopped, or from rest to a final velocity.

The "a" in the equation refers to the ACCELERATION. It is the change in velocity per unit time. But there is a catch on this variable. We often see the problems using "g" as "a". This "g" is the acceleration to due gravity, the acceleration that the object acquired because of the gravity. That is, the object in motion is either going up or down.

So now we solve the problems.

1. Jay drops a stone from rest at the top of 55m tall building.With what Velocity does it hit the ground?

The keyword here in the problem is the word "drops". So basically the problem talks about a FREE FALLING BODY (which is the motion at which the body is initially at rest, or simply the body is being dropped meaning initial velocity is zero (0).)

Since the problem asks for the final velocity and already gave the distance, we make use of formula number 3.

(Vf)^2 = (Vi)^2 + 2ad

Substituting the given in the formula, we have

(Vf)^2 = (Vi)^2 + 2ad

(Vf)^2 = (0)^2 + 2(9.81m/s^2)(55m)

(Vf)^2 = 0 + 1,079.1 m^2/s^2

Getting the square root of both sides, we have

Vf = 32.85 m/s

Therefore, the final velocity is 32.85 m/s

2. A ball drops freely from the roof towards the ground . What is the distance of the roof from the ground if the time lapsed when the ball hits the ground is 8 seconds?

This problem is also a free fall problem because the ball in the problem is "being dropped".

Base on the given, we make use of formula 1

d = (Vi)(t) + (1/2)(a)(t^2)

Substituting we have

d = (0)(8s) + (1/2)(9.81 m/s^2)((8s)^2)

d = (0) + (1/2)(627.84m)

d = 313.92 m

Therefore, the distance of the roof from the ground is 313.92m.

Last we have problem 3.

3. Â If a ball falls, From the roof top of a building towards the ground in 10s, How tall is the building?

The problem again is a free fall, and base on the given we make use of formula 3 again.

d = (Vi)(t) + (1/2)(a)(t^2)

d = (0)(10s) + (1/2)(9.81 m/s^2)((10s)^2)

d = 0 + (1/2)(981m)

d = 490.5 m

Thus, the building is 490.5 meters tall.

For more related problems, see links below.