Calculate Total Fall Time with Kinematic Equations | FreeFall Question

[cmsoccer21]

Homework Statement

A rock is thrown horizontally from a cliff with a speed of 15 m/s. It falls half of the heigh of the cliff in the last three seconds of its fall. What is the total fall time?

Homework Equations

The Kinematic Equations

The Attempt at a Solution

Well here's what I have so far but I don't know what its getting me.
Vi=Initial Velocity
Vo=Halfway Down Cliff Velocity
Vf=Final Velocity
Y=Height of cliff
t=time
Vf+29.4=-9.8t
Y=.5(0+Vo)t
Y=.5(Vo+Vf)3
Y=-4.9t^2
Y=Vo(-88.2)

These were all derived from kinematic equations.

 

[CompuChip]

I'd (always) work from the general formula,
$$y = -\frac{1}{2} g t^2 + v_0 t + h_0$$.

You have one piece of information that you can plug into that formula, to replace the unknown height $h_0$ of the cliff by the falling time $t_f$. (You know $v_0$, but think before you write it down!) Then the remaining formula only contains the constant $g \approx 9,8$ and the falling time $t_f$ so you can set it to the final height (0) and solve for the latter.

 

[baba89]

However, to calculate the total fall time, we need to use the equation: Y = Vi*t + 0.5*a*t^2, where Y is the height of the cliff, Vi is the initial velocity (15 m/s in this case), a is the acceleration due to gravity (-9.8 m/s^2), and t is the total fall time.

We can rearrange this equation to solve for t:

t = (-Vi ± √(Vi^2 + 2*a*Y)) / a

Substituting the values given in the problem, we get:

t = (-15 ± √(15^2 + 2*(-9.8)*(0.5*Y))) / (-9.8)

t = (-15 ± √(225 + (-9.8*Y))) / (-9.8)

Since we know that the rock falls half of the height of the cliff in the last three seconds, we can set Y = 0.5*Y and solve for t:

3 = (-15 ± √(225 + (-9.8*0.5*Y))) / (-9.8)

Solving for Y, we get Y = 29.4 meters.

Plugging this back into our equation for t, we get:

t = (-15 ± √(225 + (-9.8*29.4))) / (-9.8)

t ≈ 3.56 seconds

Therefore, the total fall time of the rock is approximately 3.56 seconds.