Question

A ball is droped from a building with 40
m see initial speed Height of building is
45 mts and after T seconds distance
between ball and ground level is S = 48 +40t-8t^2then after how many seconds ball
reach the ground​

Answer 1

Hey mate !!!

Your reference to speed in the question is not clear.

Yet I'm trying to solve by assumption.

Acc/to data ;

The distance between ball and Ground is given as S= 48+40t-8t^2

By differentiating it , we'll get the velocity "v"

Therefore ;

dS/dt = 40-16t

=> v= 40 -16t

as we know at the time of reaching the ground , the velocity will become zero ( 0 )

so ;

40-16t= 0

=> 16t = 40

=> t = 2.5 seconds.

Hope it helps!!!

please let me know if there's any error.