Question

you are walking on a bridge and accidentally drop your cell phone. If the bridge is 61.3 m above the water, how long does it take the phone to hit the water?

Answer 1

Answer: 3.54s

Explanation:

Answer 2

The time of flight is 3.54 s

Explanation:

The motion of the cell phone is a free fall motion (constant acceleration, $g=9.8 m/s^2$, towards the ground), since it is acted upon the force of gravity only: therefore, we can use the following suvat equation

$s=ut+\frac{1}{2}at^2$

where:

s = 61.3 m is the vertical displacement of the phone

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration

t is the time of flight of the phone

Solving for t, we find:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(61.3)}{9.8}}=3.54 s$

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