# ncert eg-3.4 motion in one dimension

Example 3.4
A ball is thrown vertically upwards with a velocity of 20 m sâ€“1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how long will it be before the ball hits the ground? Take g = 10 m sâ€“2.

given in solution (2nd method in book):

1. The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen..
2. we do not have to worry about the path of the motion as the motion is under constant acceleration..

I didnt understnd this method. please explain me.

62
Lokesh Verma ·

See the whole formula or method that is used is that

s=ut+1/2at2

Now these points s, u, a are actual position vectors (in a linear case, they are all like points a real line!)

So S will be the net displacement. Here, it will be -25 (If you take the upwards direction as +ve!)

Similarly, U will be +ve upwards=20m/s

and a=-10m/s2

Did u get the whole reason of your 2nd point and the method employed?