Question

Against the park ranger's advice, a visitor at a national park throws a stone horizontally off the edge of a 83 m high cliff and it lands a distance of 93 m from the edge of the cliff, narrowly missing a visitor below.
What was the initial horizontal velocity of the rock, in m/s? You can round your answer to the hundredths place and use g ≈ 10 m/s2.

Answer 1

Answer:

20.37m/s

Explanation:

For this problem, we'll need a basic kinematic equation for both directions

Vertical direction:

$y(t)=\frac{1}{2}a_{y}t^{2}+v_{o_y}t+y_{o}$

Knowing that the acceleration in the y-direction is due to gravity, and is constant near the surface of the Earth, where the rock is being thrown, we get $a_{y}=g=-10m/s^{2}$.  We're also given that the initial vertical height is 83m, so $y_o=83m$.  Additionally, we're given that the stone is thrown horizontally, and thus initially, it has no vertical velocity.  Only velocity in the x-direction.  Thus, $v_{o_y}=0m/s$.  Substituting known values into our equation:

$y(t)=\frac{1}{2}(-10)t^{2}+(0)t+(83)$

$y(t)=-5t^{2}+83$

This equation is true for all times until the acceleration changes (isn't constant... i.e. when it hits the ground).

What now?

We want to find the specific time when it does hit the ground.  Why?  Because this is the point in time where we know some extra information about what is happening in the x direction, and that will help us deal with the x-direction equation.  

When the stone does hit the ground, the output of this function (measuring the height of the stone) is zero, which leaves only one unknown value, t... the time associated with that y:  the time that it takes to hit the ground:

$0=-5(t_{ground})^{2}+83\\-83=-5(t_{ground})^{2}\\\frac{-83}{-5}=(t_{ground})^{2}\\16.6=(t_{ground})^{2}\\\sqrt{16.6}=\sqrt{(t_{ground})^2}\\4.074309757seconds=t_{ground}$

Again, this is the time that the stone hits the ground.  Knowing how special this time is, one might give it a special subscript to keep track of it.  

$t_{ground}=4.074309757seconds$

Horizontal direction

Reusing the basic kinematic equation for the horizontal direction:

$x(t)=\frac{1}{2}a_{x}t^{2}+v_{o_x}t+x_{o}$

Note that in the horizontal direction, there is no acceleration (after the rock is released), so $a_{x}=0$.  Also, if we define the base of the cliff as our origin, then at the top of the cliff where the rock is released, the initial x-position is still zero, aka, $x_{o}=0$

$x(t)=\frac{1}{2}(0)t^{2}+v_{o_x}t+(0)\\x(t)=v_{o_x}t$

While we don't know the initial horizontal velocity, we do know an output of the function for one specific input... we know that when the stone hits the ground, that the stone's horizontal distance is 93m from the cliff.  So, if we input $t_{ground}$ into this function, it will give an output of 93.  The only remaining unknown is the initial velocity in the x direction.  One equation, one unknown, we can solve this:

$x(t_{ground})=v_{o_x}t_{ground}\\83=v_{o_x}(4.074309757)\\\frac{83}{4.074309757} =v_{o_x}\\20.37154879m/s =v_{o_x}$

If there were a component of initial velocity in the y direction, we would need to use the Pythagorean theorem to combine them.  However, as the initial y component of velocity is zero, the initial velocity is simply the initial x velocity.

Paying attention to directions, round to the answer to the nearest hundredth:

$20.37m/s =v_{o}$