Question

A ball thrown up is caught back by the thrower , what is the net displacement of the ball

Answer 1

There is a formula for that..

S= displacement

Sn = displacement at nth second

Sn=u+a/2(2n-1)

S=ut+1/2at²

Now time of flight is = (2u sinx)/g

Sinx =1 as thrown upwards…

Now time of flight is 4 sec

4= 2u/g. Assuming g =9.8m/sec²

U=19.6m/sec

Now put this value of u in he above equation

g=-9.8m/sec² as it acts downwards…and. We to-ok the velocity upwards +ve here .

Just we know that maximum height is attained at 2 sec ,half of time of flight..

So assuming that to be 0 on displacement line, displacement by the ball in the 2→3 sec is

U=0 , as velocity becomes 0 at maximum height.

S from that

S= 0*1+1/2 (9.8)1²

S= 4.9m

Another conventional methid to solve this easy sum

S=ut+1/2at²

S= 19.6*3 +1/2 (-9.8) t²

S= 58.8–(4.9*9)

58.8–44.1= 14.7 from the bottom

Now..

Time of flight is 4 sec..

Max. Height is at 2 sec

S@2 sec , =19.6*2 +1/2*(-9.8)t²

39.2–19.6= 19.6

Therefore

19.6–14.7= 4.9 metre from the top