Question

A blue ball is thrown upward with an initial speed of 24.3 m/s, from a height of 0.9 meters above the ground. 3 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 8.2 m/s from a height of 33.3 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s².

4) What is the height of the red ball 3.9 seconds after the blue ball is thrown?

5) How long after the blue ball is thrown are the two balls in the air at the same height?

Answer 1

Let us write the general solution first. Suppose, a ball is thrown at time t=0 from height y(0), in the upward direction with a constant velocity u and you want to know its height at a later time t=t. What would be the equation of motion?

We have only one direction. Lets call it y axis (upward is positive direction). The gravitational acceleration g is acting along the negative y axis. Hence, the equation of motion would be,

Another integration with respect to t gives,

4) Now get back to the case of the red ball first. For this ball we have, y(0) = 33.3 m, m/s, (-ve because it is thrown along the negative y axis). You want to know the height of the red ball 3.9 seconds after the blue ball is thrown i.e. 0.9 seconds after the red ball was thrown. The answer is, y(0.9), which we can calculate from the boxed equation,

5) Suppose those two balls will be at same height at time t after the blue ball is thrown. Then the height of the blue ball will be,

At that time the red ball will be at a height,

As they are at same height, these two are same and we get,

Now put the numerical values to get,