You are driving off of a horizontal cliff, and your car has a horizontal velocity of 29.33 m/s at the instant you leave the cliff. The cliff is an unknown height above a rocky ravine, but you land 146.8 m away from the sheer cliff face.
How tall is the cliff?
Horizontal velocity does not change in a free-fall, so
time = distance/velocity
= 146.8/29.33
= 5.0051 s. (keeping 5th significant fig.)
Initial vertical velocity = 0 for free fall, so
H=(1/2)at^2
=(1/2)*(-9.81)(5.0051^2)
=-122.9 m
Ans. the ground is 122.9m below the cliff.
To find the height of the cliff, we can use the equations of motion. Let's assume the initial vertical position of the car when it leaves the cliff as zero.
We are given:
Initial horizontal velocity (Vx) = 29.33 m/s.
Horizontal displacement (Δx) = 146.8 m.
Let's consider the vertical motion of the car. The only force acting vertically is gravity (acceleration due to gravity, g = 9.8 m/s^2) in the downward direction. The vertical motion can be described using the equation for displacement:
Δy = Vyi * t + (1/2) * g * t^2
Here, Δy represents the vertical displacement (height), Vyi represents the initial vertical velocity (which is zero in our case), t represents the time of flight, and g represents the acceleration due to gravity.
Since the vertical displacement is the height of the cliff, we can substitute the values we know:
Δy = 0 + (1/2) * 9.8 * t^2
Δy = (4.9) * t^2
Now, let's consider the horizontal motion of the car. We know the horizontal velocity of the car remains unchanged throughout its flight. The horizontal displacement can be calculated using the equation:
Δx = Vx * t
Substituting the values we know:
146.8 = 29.33 * t
Now we have two equations:
Δy = 4.9 * t^2
Δx = 29.33 * t
We can solve the second equation for time (t) and substitute it into the first equation to find the height of the cliff (Δy). Let's do that:
t = Δx / Vx
t = 146.8 / 29.33
t ≈ 5 seconds (rounded to two decimal places)
Now we can substitute the calculated time (t) into the first equation:
Δy = 4.9 * (5)^2
Δy = 4.9 * 25
Δy = 122.5 meters
Therefore, the height of the cliff is approximately 122.5 meters.