A ball is dropped from rest from the top of a building of height ?h?. At the same instant, a second ball is projected vertically upward from ground level, such that it has zero speed when it reaches the top of the building. When the two balls pass each other, which ball has the greater speed, or do they have the same speed? Explain. Where will the two balls he when they are alongside each other: at height ?h/?2 above the ground below this height, or above this height? Explain.
Solution 20DQ Let us have a look at the following figure. Let us assume that the two balls meet at a distance of x from the top as shown in the figure above. For the first ball, initial speed is zero. Let1v be the speed of the ball at the distance x. So, v 2 = 2gx 1 v = 2gx …..(1) 1 Let the second ball be projected with a speed of u 2 and its speed at the height h x be v 2 Therefore, v 2 = u 2 2g(h x) 2 2 v 2 = u 2 2gh + 2gx…..(2) 2 2 Now, at the top height for the 2nd ball, final velocity is zero. 2 So, u 2 2gh = 0 Therefore, from equation (2) v 2 2= 2gx v 2 2g …..(3) Therefore, from equations (1) and (3), both the balls will have the same speed at the height x. The two balls will meet above the h/2 height from the ground.This is because the first ball is initially at rest and so, it will get the downward velocity very slow. The second ball is already thrown with an initial velocity. So, it will move faster the first half of the height than the first ball.
