Wile E Coyote, having missed the Road Runner, leaves the edge of the cliff at 29 m/s horizontal velocity. If the canyon is 104 m deep, how far from the base of the cliff does the coyote land?
time to fall: h=1/2 g t^2 solve for time t.
how for horizontal? d=vi*time
To solve this problem, we can use the equations of motion to analyze the coyote's horizontal and vertical motion separately.
Let's start by finding the time it takes for the coyote to fall to the base of the cliff using the vertical motion equation:
Δy = Viy * t + 0.5 * g * t^2
Where:
Δy is the vertical displacement (104 m),
Viy is the initial vertical velocity (0 m/s),
t is the time,
g is the acceleration due to gravity (-9.8 m/s^2).
Since the coyote starts from rest vertically (Viy = 0 m/s), the equation becomes:
Δy = 0.5 * g * t^2
Simplifying the equation, we have:
104 = 0.5 * 9.8 * t^2
Solving for t^2:
t^2 = (2 * 104) / 9.8
t^2 = 21.2244898
Taking the square root of both sides:
t ≈ 4.6087 seconds
Now that we have the time it takes for the coyote to fall to the base, we can find the horizontal displacement using the horizontal motion equation:
Δx = Vix * t
Where:
Δx is the horizontal displacement,
Vix is the initial horizontal velocity (29 m/s),
t is the time.
Substituting the given values:
Δx = 29 * 4.6087
Δx ≈ 133.6983 meters
Therefore, the coyote lands approximately 133.6983 meters from the base of the cliff.