Answers:
a) 12.54 m
b) 1.6 s
c) 0 m/s
Explanation:
This problem can be solved with the following equations:
[tex]y=y_{o}+V_{o}sin \theta t +\frac{1}{2}gt^{2}[/tex] (1)
[tex]V_{f}=V_{o}+gt[/tex] (2)
Where:
[tex]y=0m[/tex] is the phone's final height
[tex]y_{o}[/tex] is the phone's intial height and the cliff's height
[tex]V_{o}=0 m/s[/tex] is the initial velocite, since the phone was dropped
[tex]t[/tex] is the time it took to the phone to hit the water
[tex]g=-9.8 m/s^{2}[/tex] is the acceleration due gravity, always directed downwards
[tex]\theta=90\°[/tex] is the angle with the horizontal
[tex]V_{f}=-15.75 m/s[/tex] is the phone's final velocity, directed downwards
a) Let's begin by rewriting (1) and (2) acording to the information given above:
[tex]y_{o}=-\frac{1}{2}gt^{2}[/tex] (3)
[tex]V_{f}=gt[/tex] (4)
Isolating [tex]t[/tex] from (4) and substituting in (3):
[tex]y_{o}=-\frac{1}{2}g(\frac{V_{f}}{g})^{2}[/tex] (5)
[tex]y_{o}=-\frac{1}{2}-9.8 m/s^{2}(\frac{-15.75 m/s}{-9.8 m/s^{2}})^{2}[/tex] (6)
[tex]y_{o}=12.54 m[/tex] (7) This is the height of the cliff
b) Finding [tex]t[/tex] from (4):
[tex]t=\frac{V_{f}}{g}[/tex] (8)
[tex]t=\frac{-15.75 m/s}{-9.8 m/s^{2}}[/tex] (9)
[tex]t=1.6 s[/tex] (10)
c) Since [tex]\theta=90\°[/tex] the horizontal velocity is:
[tex]V_{x}=V cos (90\°)=0 m/s[/tex]
