Question

A ball is thrown straight down with an initial speed of 14.7 M per second from the the top of building 49 M high. find the time at which i it reaches the ground below

Answer 1

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Given :

Initial speed =u= 14.7m/s

height =h=49mts

t=?

acceleration due to gravity =g=9.8m/s^2

v=0m/s ..since it reaches the ground after being thrown


Now,

v=u+at

0=14.7+9.8t

14.7=9.8t

t=14.7/9.8

=1.5 sec...


I'm not sure