If you are standing on a cliff that is 160 ft high, how long will it take a rock to hit the ground if you drop it?

Question

I know we need to use h=-16t^2+v(0)t+h(0). But I'm not exactly sure how to put it into the equation properly

Answer 1

This is a physics and pre-calculus problem.

 

These are the formulae that you need to know to solve any problems about object in motion:

 

Vf = Vi + gt

 

S = (1/2)gt²+ Vi*t

 

Vf² =Vi² + 2gS

 

Where:

S = distance or height

Vf= final velocity

Vi = initial velocity

g = acceleration (due to gravity = 32 ft/sec², 9.8 m/sec², 980 cm/sec²).

t = time lapsed

 

In your problem's case, the rock was dropped therefore, the initial velocity is zero (Vi=0).

I see that you are using the 32 ft./sec² as your acceleration due to gravity because you use 16t² where in the second formula shown above is gt². However you can't use negative 16t² (-16t²) because if the object goes down, it's going faster and faster as the time goes by. You can only use positive 16t² if the rock is thrown upward.

 

Don't get confuse with velocity.  The velocity of the object going up is positive and if it is going down then the velocity is negative. To remember this, think of the y-axis (vertical axis) of the coordinate plane.

 

Therefore,

 

S= 160

g = 32

Vi = 0

t = ?

 

 

Use the second formula above.

S = (1/2)gt²+ Vi*t

160 =(1/2)(32) t² + (0)t

160 = 16t^2

10 = t^2

 

t = sqrt (10) = approx. 3.16 seconds

Answer 2

Hi Taylor, using the equation, h = -16t² + v₀ * t + h_o, we can solve for t.

 

Constraints:

 

v₀ * t = 0, there is no initial velocity.

 

h₀ = 160 feet.

 

h = 0, it hits the ground

 

Solve for t:

 

0 = -16t² + 0 + 160

 

Subtract 160 from both sides:

 

-160 = -16t²

 

Divide both sides by -16 and rearrange:

 

t² = 10

 

t = ±√10

 

t ≈ ± 3.16

 

The time it takes to hit the ground is: t ≈ 3.16 seconds.

 

We can discard the solution: t ≈ -3.16 seconds, since the time it takes to fall will be positive.

 

Questions?

 

 

Answer 3

I'm not sure where you're getting the 16 from in your equation, but this problem relies on the acceleration due to gravity, a=9.18m/s^2.  I like a little different form of the distance formula, x=x₀ + v₀t + 1/2a₀t².  Then you just plug in the numbers you know, and solve for the rest.

 

The little catch here is that 160ft isn't in meters, which is presumably the unit you want, so you should convert it.  1 foot is about .3048 meters, so 160x.3048=48.768, and you can tailor that to whatever significant figures.

 

You already know that there is no initial velocity in this problem, since you're dropping the rock (from rest), rather than throwing it.  Your equation should now look like this:  0=48.768m + 0t + -(1/2)9.8m/s²(t²).  The reason the acceleration here is negative is just for convenience in establishing directions - your height is positive, so falling from that height is negative.

 

Now you just drop the unnecessary stuff from your equation and solve for t, so your next step is to add (1/2)9.8m/s²(t²s²) to both sides, and get to here:  (1/2)9.8m/s²(t²)=48.768m.  Divide both sides by 4.9m/s² to get t²=9.953/s², then take the square root:  sqrt(t²)=sqrt(9.953/s²)=3.15s, or t = 3.15 seconds, meaning that the rock takes 3.15 seconds to hit the ground from a 160 foot high cliff.

 

Edit:  I see now that the 16 is from 1/2at², with acceleration due to gravity being 32f/s².  I forgot the 1/2 in my equations originally, sorry!