Answer :
Answer:
19.81m/s
Explanation:
To find the speed at which the car leaves the cliff, first we need to know how long it took to reach the ground. We do this with the equation:
[tex]y=y_{0}-\frac{1}{2}gt^2[/tex]
Where [tex]y[/tex] is the vertical position at a time [tex]t[/tex], since we are looking for the moment when the car reaches the gound: [tex]y=0[/tex]. [tex]y_{0}[/tex] is the initial vertical position, in this case the height of the cliff: [tex]y_{0}=125m[/tex]. [tex]g[/tex] is gravitational acceleration: [tex]g=9.81m/s^2[/tex].
So replacing the known values:
[tex]0=125-\frac{1}{2}(9.81m/s^2)t^2[/tex]
and since we need the time, we clear for it:
[tex]0=125-\frac{1}{2}(9.81m/s^2)t^2\\-125=-\frac{1}{2}(9.81m/s^2)t^2\\(-125)(-2)=(9.81m/s^2)t^2\\\frac{250}{9.81m/s^2}=t^2 \\25.48=t^2\\\sqrt{25.48}=t\\ 5.048=t[/tex]
At time [tex]t=5.048[/tex] the car reaches the gound, according to the problem at a horizontal distance from the base of the cliff of 100m.
To find the velocity we use:
[tex]x=x_{0}+v_{0}t[/tex]
where [tex]x[/tex] is the horizontal distance at time [tex]t[/tex] (in this case [tex]x=100[/tex] and [tex]t=5.048[/tex]), [tex]x_{0}[/tex] is the initial horizontal distance, we define the cliff as zero in horizontal distance so [tex]x_{0}=0[/tex], and [tex]v_{0}[/tex] is the velocity of the car when it rolled of the cliff.
replacing the known values:
[tex]100=0+v_{0}(5.048)[/tex]
and we clear for [tex]v_{0}[/tex]:
[tex]\frac{100}{5.48} =v_{0}\\19.81m/s=v_{0}[/tex]
The initial velocity (when it rolled of the cliff) is 19.81m/s