initially coin is at rest and then it drop for total time t = 1.5 s
so here the speed of the coin at which it will hit the floor is to be find
[tex]v_f = v_i + at[/tex]
here we know that
[tex]v_i = 0[/tex]
a = 9.8 m/s^2
t = 1.5 s
now from above equation
[tex]v_f = 0 + 1.5 (9.8)[/tex]
[tex]v_f = 14.7 m/s[/tex]
so it will hit the floor with speed 14.7 m/s
The speed of coin as it hits the ground is 14.7 m/s.
Given data:
Initial speed of coin is, [tex]u=0 \;\rm m/s[/tex].
Time taken to fall on ground is, [tex]t = 1.5 \;\rm s[/tex].
Use first kinematic equation of motion as,
[tex]v = u+gt[/tex]
here, g is the gravitational acceleration.
Solving as,
[tex]v = 0+9.8 \times 1.5\\v= 14.7 \;\rm m/s[/tex]
Thus, the speed of coin as it hits the ground is 14.7 m/s.
Learn more about kinematic equation of motion here:
https://brainly.com/question/13202578?referrer=searchResults
