Question

1. Olivia is taking pictures of the ocean at the edge of the cliff when she accidently drops her phone. The phone hit the water at a vertical

velocity of 15.75m/s and landed 16m away from the base of the cliff.
a. How tall is the cliff?
b. How long did it take for the phone to hit the water?
C. What was the phone's horizontal velocity?

Answer 1

Answers:

a) 12.54 m

b) 1.6 s

c) 0 m/s

Step-by-step explanation:

This problem can be solved with the following equations:

`y=y_(o)+V_(o)sin \theta t +(1)/(2)gt^(2)` (1)

`V_(f)=V_(o)+gt` (2)

Where:

`y=0m` is the phone's final height

`y_(o)` is the phone's intial height and the cliff's height

`V_(o)=0 m/s` is the initial velocite, since the phone was dropped

`t` is the time it took to the phone to hit the water

`g=-9.8 m/s^(2)` is the acceleration due gravity, always directed downwards

`\theta=90\°` is the angle with the horizontal

`V_(f)=-15.75 m/s` is the phone's final velocity, directed downwards

a) Let's begin by rewriting (1) and (2) acording to the information given above:

`y_(o)=-(1)/(2)gt^(2)` (3)

`V_(f)=gt` (4)

Isolating
`t` from (4) and substituting in (3):

`y_(o)=-(1)/(2)g((V_(f))/(g))^(2)` (5)

`y_(o)=-(1)/(2)-9.8 m/s^(2)((-15.75 m/s)/(-9.8 m/s^(2)))^(2)` (6)

`y_(o)=12.54 m` (7) This is the height of the cliff

b) Finding
`t` from (4):

`t=(V_(f))/(g)` (8)

`t=(-15.75 m/s)/(-9.8 m/s^(2))` (9)

`t=1.6 s` (10)

c) Since
`\theta=90\°` the horizontal velocity is:

`V_(x)=V cos (90\°)=0 m/s`