Question

A ball is dropped from a height of 49m. Find the time interval in which the ball hits the ground. (g=9.8 m)

Plz show full method.....

Answer 1

Hi. ..


Given :
S= 49m
a = 9.8ms^-2
u = 0ms ^-1.

Thus by using .

S= ut + 0.5 at^-2

49 = 0 + 9.8 * (t)^2

Thus 49 / 9.8 = (t)^2

Thus t = 2.3 sec..

Hope this helps u!!

Answer 2

HERE WE HAVE, 
                   (U) INITIAL VELOCITY = 0 m/s
                    (S) DISPLACEMENT = 49 m
                    (a) acceleration = 9.8 m/ s^2
now, 
        by applying formula, 
        s= ut + ({at^2)/2
       => 49 = 0 + {(9.8* t^2) / 2}
       => 98 = 9.8 * t^2
       => 10 = t^2
   therefore time  = √10 = 3.16 seconds