Question

A ball is projected uowards with certain velocity. While moving upwards ot passes through point a in t1 seconds and comes on the ground in t1+t2 seconds. Height of a abive the point of projection is

Answer 1

Answer:

Explanation:

Let initial projected speed pf ball be ‘u’ vertically up.

So in time t1

h = ut1 – ½ gt1 2    ( we have to find h )......................................................................(1)

 

We ball returns to the ground the displacement is ZERO.

S = ut +1/2 gt2

0 = u(t1 ​+ t2) – ½ g (t1 ​+ t2)2

Solvinng quadratic,

t1 + t2  = 0  or 2u/g  (neglectinng zero as it is the initial time)

t1 ​+ t2 = 2u/g...............................................(2)

 

 

Substitute ‘u’  from 2 in 1

h =1/2 gt1t2