Learning Objectives To be familiar with the procedures of so

Learning Objectives To be familiar with the procedures of solving science problems using computing. To be able to implement science formulas in computer programs. To be able to analyze and evaluate the computing results Background You are about to solve problems of straight-line motion with constant acceleration. The problems are typically studied in high school physics. Some useful formulas are given below: v = v_0 + at x = x_0 + v_0 t + 1/2 at^2 v^2 = v^2_0 + 2a(x - x_0) where v is velocity at time t; v_0 is initial velocity at time t = 0; x is position at time t; x_0 is initial position at time t = 0; a is acceleration. In free fall case, the acceleration due to gravity is a constant, g = 9.81m/s^2 and always downward. Assignments Use the formulas above to solve the following problems. A ball is dropped from a high building. It starts from rest and falls freely. Calculate its position (distance from the ground) and velocity after 0.2s, 0.4s, and 0.6s. If the building is 10m high, what is the velocity of the ball when hitting the ground? After hitting the ground, the ball bounces back upward at 70% of the velocity at which the ball hits ground due to the energy loss during collision between the ball and the ground. How high would the ball bounce up? How long does it, take for the ball to fall back to the ground again?

Solution

Answer as follow to your questions with respective question number:

Answer 1:

We know that: v = v0 + at

and  x=x0 + v0t + 1/2at^2

tha ball starts from rest so initial velocity v0=0 and initial distance x0=0

substitute above values in  v = v0 + at and x=x0 + v0t + 1/2at^2 we get

v = at and here a=9.81 m/s^2 as ball falls freely

x=1/2at^2

So,

We can calculate velocities at respective time as follows using v = at formula :

therefore velocity after 0.2sec is v = at

=9.81 * 0.2

v = 1.962 m/s

again velocity after 0.4sec is v = at

=9.81 * 0.4

v = 3.924 m/s

velocity after 0.6sec is v = at

=9.81 * 0.6

v = 5.886 m/s

Now calculating respective ball position at given time using above formula  x = 1/2at^2

so, position x at 0.2 sec is x = 1/2 at^2

=1/2 * 9.81 * 0.2^2

=0.1962m

position x at 0.4 sec is x = 1/2 at^2

=1/2 * 9.81 * 0.4^2

=0.7848m

position x at 0.6 sec is x = 1/2 at^2

=1/2 * 9.81 * 0.6^2

=1.7658m

For any queries revert back.Thanks and regards.