Answer:
The speed was 5.42 m/s
Explanation:
The height of fall of the phone s = 1.5 m
The phone was at rest in her hand, so the initial velocity u = 0 m/s
The velocity right before the phone strikes the ground v = ?
Using the equation
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2gs
where
g is the acceleration due to gravity = 9.81 m/s^2
substituting values, we have
[tex]v^{2}[/tex] = [tex]0^{2}[/tex] + 2(9.81 x 1.5)
[tex]v^{2}[/tex] = 29.43
v = [tex]\sqrt{29.43}[/tex] = 5.42 m/s
