Answer :
Answer:
a) [tex]t=0.60 s[/tex]
b) [tex]v_{ox} =0.5m/s[/tex]
Explanation:
From the exercise we have initial height and final X position
[tex]y_{o}=1.807m[/tex]
[tex]X=0.3012m[/tex]
a) From the concept of free falling objects we have that
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
Since the car runs off the edge of the table, that means the car is moving in x direction with [tex]v_{oy}=0m/s[/tex] and at the end of the motion [tex]y=0m[/tex]
[tex]0=1.807m-\frac{1}{2}(9.8)t^{2}[/tex]
Solving for t
t=± 0.6072 s
Since the time can not be negative, the answer is t=0.6072 s
b) To find the horizontal velocity of the car, we need to use the time that we just calculate
[tex]X=v_{ox}t[/tex]
[tex]v_{ox}=\frac{X}{t}=\frac{0.3012m}{0.6072s} =0.5m/s[/tex]