Answer: 0.522 m
Explanation:
This situation is related to projectile motion, where the initial velocity of the car [tex]V_{o}=24.5 m/s[/tex] has only the horizontal component. This means the angle is equal to zero ([tex]\theta=0[/tex]). And the equations we will use to find the height [tex]y_{o}[/tex] of the cliff are:
[tex]y=y_{o}+V_{o}sen \theta t+\frac{1}{2}gt^{2}[/tex] (1)
Where:
[tex]y=0 m[/tex] is the final height of the car
[tex]y_{o}[/tex] is the initial height
[tex]V_{o}=24.5 m/s[/tex] is the initial velocity of the car
[tex]t[/tex] is the time
[tex]g=-9.8 m/s^{2}[/tex] is the acceleration due to gravity
[tex]\theta=0[/tex] is the angle
[tex]x=8 m[/tex] is the horizontal distance traveled by the car after passing the cliff
Well, firstly we have to find [tex]t[/tex] from (1):
[tex]0=y_{o}+\frac{1}{2}gt^{2}[/tex] (3)
[tex]t=\sqrt{-\frac{2y_{o}}{g}}[/tex] (4)
Substituting (4) in (2):
[tex]x=V_{o}cos \theta (\sqrt{-\frac{2y_{o}}{g}})[/tex] (5)
Isolating [tex]y_{o}[/tex]:
[tex]y_{o}=-\frac{x^{2} g}{2V_{o}^{2}}[/tex] (6)
[tex]y_{o}=-\frac{(8m)^{2} (-9.8 m/s^{2})}{2(24.5)^{2}}[/tex] (7)
Finally:
[tex]y_{o}=0.522 m[/tex]
