Question

an object is dropped from the top of the tower of height 160m and at the same time another object is thrown vertically upward with the velocity 80m/s from the foot of tower.state that when and where the object meet ​

Answer 1

Answer:

140.4 m

Explanation:

Since both are subjected to the same g , we can ignore g and do the problem.

The ball at the top is at rest , ( no g) . The thrown up ball moves up with uniform speed of 80 m/s , to meet the ball at the topn( no g)

Time to go up a distance of 160 m is 160/80 = 2 s .

In 2 second it meets the other.

To find the place of meet in the real situation is

S = 1/2 g t^2 = 4.9 * 4 = 19.6 m down from the tower

Or 160 - 19.6 = 140.4 m up from the ground