Question

A ball thrown vartically upwords with the speed 9.8m/s how high will the ball rise and how long will it rise​

Answer 1

Answer:

Let X be the meeting point.

Both balls would have traveled same time to get to meeting point, say 't' secs.

Let h 1 be height traveled by ball falling down to X.

Let h  2  be height traveled by ball going up to X and v be its velocity.

h  1+h 2

​ =100m

v  2=2gH=2×9.8×100=1960

Hencev=  1960  

m/s  h  1  

=  2 1  gt  2  

For the ball travelling up,

h  2  =vt−  21

gt  2  =vt−h  

1 h  

1 +h  +2  =vt

Hence vt=100

t=  v 100

​ =1960

100  secs.

 h  1 =  2 1

​ gt  2  =  21  ×9.8×  

1960

100×100

​ =  4 100 ​ =25m

h=  2  =100−25=75m

Explanation: