kinematic motion

Question

Write the kinematic equation for distance as a function of initial speed (v0), constant acceleration (a) and time (t)

A ball is thrown straight up in the air from the ground with an initial speed of V0. Write an equation that shows the vertical position of the ball Y(t) as a function of time.

For a building having a height equal to the six times the quantity you have recorded for Day in meters (in our example of July 15, this would be 6x15 = 90 meters), compute the time required for the ball to fall to the ground while experiencing acceleration due to gravity (g=9.8m/s2). How fast was the ball traveling when it hit the ground? Show all work.

Answer 1

The equation for a body that moves with constant acceleration is d(t) = d0 + v0 t + 1/2 a t^2 where d0 is the initial position of the body, v0 is the initial velocity and a is the acceleration.

For a ball under the acceleration of gravity g, the height of the body at a time t is given by Y(t) = Y0 + v0 t - 1/2 g t^2 where the minus sign in front of g is because we are assuming the positive direction of the y coordinate to be up (and negative down).

For the third problem, we put into practice everything we learned so far.

In this case, we have Y(t*) = 0 m since the ball touches the ground at the final instant, Y0 = 90 m, v0 = 0 m/s since the ball is left to fall and the gravity acceleration g is directed towards down. In this case we have: Y(t*) = Y0 + v0 t* -1/2 g t*^2 -> Y0 = 1/2g t*^2 -> t* = √(2Y0/g) = 4.28s.

Finally, the final velocity of the ball v(t*) at the final time t* is given by the equation v(t*) = v0 - g t* = -g t*-> v(t*) = - 42.02 m/s which means the velocity is directed towards the ground as we expect