Question

You are walking on a bridge and accidentally drop your cell phone. If the bridge is 45.4 m above the water, how long does it take the phone to hit the water?

Answer 1

3.04 seconds

Step-by-step explanation:

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Answer 2

The phone is in free fall, so it is moving by uniformly accelerated motion, with constant acceleration equal to
`g=9.81 m/s^2` (gravitational acceleration). Taking the water as reference point, the vertical position of the phone at the istant t is

`y(t) = h - (1)/(2)gt^2`
where h=45.4 m is the height of the bridge. The phone hits the water when y(t)=0, so by requiring this condition we find the time t after which the phone hits the water:

`h- (1)/(2)gt^2=0`

`h= (1)/(2)gt^2`

`t= \sqrt{ (2h)/(g) }= \sqrt{ (2 \cdot 45.4 m)/(9.81 m/s^2) }=3.04 s`