Free Fall Question

Question

You decide to drop an apple from rest from the observation platform in the statue of liberty’s crown. The height h from which the apple is released is 44.5 meters above the shoulders of your friend Robin H., who happens to be standing on the ground directly below you. Robin H. shoots an arrow straight up at the same instant that you release the apple. The initial velocity of the arrow is 24.3 m/s.

 

a) How long after you drop the apple will the arrow hit it? b) what is the speed of the arrow right before it strikes the apple?

Answer 1

Meg,

 

Since we don't know how tall Robin H. is let measure our distance d as the distance the apple falls from the point of release on the observation platform.  Then for the falling apple:

 

d = ½·gt²

 

The arrow being fired upward with initial speed V₀ will lose speed due to the gravitational acceleration, so for the distance the apple travels we need to write:

 

(44.5m - d) = V₀·t - ½gt²

 

Replacing distance in the 2nd equation with the term for distance in the first gives:

 

44.5 m - ½gt² = V₀·t - ½gt²

 

Notice that in solving for time the ½gt² terms cancel! 

 

then 44.5 m = V₀·t ⇒ t = (44.5 m/24.3 m/s) = 1.83 s

 

The result may seem surprising, but keep in mind that as the apple fell it was gaining speed at exactly the same rate at which the arrow was loosing speed!

 

So our answer to (a) is t = 1.83 s

 

For (b) the final velocity of the arrow is given by V_f  = V₀ - a·t

 

V_f = 24.3 m/s - (9.8 m/s²)(1.83 s) 

 

V_f '= 12.4 m/s

Answer 2

When the arrow hits the apple, together the two objects will have traveled 44.5 m which is the distance between them at the start. So we write an equation for the sum of the distances.

 

We use the formula distance = v₀t - (1/2)at² with a = -9.8m/s²

 

distance traveled by apple is (.5)(9.8m/s²)t² because initial velocity is zero

 

Write the distance traveled for the arrow, add the two distances and set it equal to zero.

 

You should get t = 1.83 seconds

Answer 3

So here we have a basic kinematics question. First lets define some of our constants;

 

h=44.5 meters

V₀=24.3 meters/second

g=9.8 meters/second²

t₀= time when the two objects hit

 

Now we can use the kinematic equations (equations of motion for an object under constant acceleration). They take the form

 

Y=Y₀+V₀t+(1/2)*a*t²

 

where Y₀ is the initial height, V₀ is the initial velocity, and a is the acceleration.

For this problem we will need to have two of these equations, since there are two objects. Since the apple has no initial velocity, the equation for the falling apple will be

 

Y_a=h-(1/2)*g*t²

 

The equation for the arrow does have an initial velocity, but its initial height is zero. So the equation for the arrow will be

 

Y_A=V₀*t-(1/2)*g*t²

 

Now we want to find out when these two objects hit. In terms of the equations, that means we should set them equal to each other. If we solve for the time, this will be the time when the two objects are at the same height.

 

Y_a=Y_A

h-(1/2)*g*t₀²=V₀t₀-(1/2)*g*t₀²

h=V₀t₀

t₀=h/V₀

 

Now we can plug in the actual numbers

 

t₀=44.5/24.3

t₀=1.83 seconds

 

What is interesting about this situation, is that the effect of gravity was not part of the final answer. This is because both objects are under the influence of gravity, so when we set the two equations equal to each other, the term -(1/2)*g*t² cancels out.

 

If we want to find the speed of the arrow at the time when they hit, t₀, we need to use a different equation. This  equation is itself a basic kinematic equation, which can be/should be memorized, however if you know some basic calculus it can be derived as follows.

 

Take the equation Y=Y₀+V₀*t+(1/2)*a*t² and find its derivative with respect to t. This will produce an equation for velocity, given by

 

V=V₀-a*t

 

In this case, the acceleration, a, is actually the acceleration due to gravity, g.

 

V=V₀-g*t

 

Now we simply plug in our found value of t, and solve for V

 

V=24.3-(9.8)*(1.83)

 

V=6.37 meters/second

 

In this calculation, the gravitational force was needed.

 

Hope this helps!

Answer 4

(a)

Say, the apple drops by x mt ,when it meets the arrow.

So, the arrow goes up by (44.5-x) Mt

 

Say the apple meets the arrow after t seconds of free fall.

 

Equation of fall motion of apple----

 

x=(1/2)*g *t^2              g= 9.8 mt/sec^2

 

t =sqrt(2x/9.8)

 

Equation of rise of arrow------

 

44.5-x=24.3t-(1/2)*9.8*t ^2

 

t =1.83 sec

From the time of free fall, the apple shall meet the arrow after 1.83sec

 

(b)

Speed of arrow before strike

 

V = 24.3-9.8*t

Substitute the value of t from above

 

V =24.3-9.8*1.83

   = 12.4 mt/sec

 

That is,the speed of arrow shall be 12.4mt/sec