Please solve using MATLAB A stone is dropped at zero velocit

Please solve using MATLAB

A stone is dropped at zero velocity from the top of a building at time...

A stone is dropped at zero velocity from the top of a building at time t = 0. The differential equation that yields the displacement x from the top of the building is (with x = 0 at t = 0) d^2x/dt^2 = g - 5V where g is the magnitude of gravitational acceleration, given as 9.8 m/s^2, and V is the downward velocity dx/dt. Calculate the displacer taking the time step as 0.5 second. Plot x and V as a function of time t.

Solution

solution:

instantaneous velocity :

Is the average velocity evaluated for a time interval that approaches zero. Thus, if an object undergoes

displacement Ax in a time At, then for that object, Ax v = instantaneous velocity = time At where the notation means

that the ratio Ax/At is to be evaluated for a time interval At that approaches zero.

Graphical Interpretations:

For motion along a straight line (the x-axis) are as follows: The instantaneous velocity of an object at a

certain time is the slope of the x-versus-t graph at that time. The instantaneous acceleration of an object at a certain

time is the slope of the v-versus-t graph at that time. For constant-velocity motion, the x-versus-t graph is a straight

line. For constant-acceleration motion, the v-versus-t graph is a straight line.

Acceleration Due To Gravity (g):

The acceleration of a body moving only under the force of gravity is g, the gravitational (or free-fall) acceleration, which is directed vertically downward.