math story problem

Question

How do I solve this question?

 

       A bird is sitting 16 feet above the ground in an appple tree dislodges an apple. After how many seconds does the apple land on the ground?  (assume no branches interfere with its fall)

 

 

Is there a factor expression that could be written to solve this?

Answer 1

"A bird is sitting 16 feet above the ground in an apple tree dislodges an apple. After how many seconds does the apple land on the ground?"

 

This is a classic falling body problem. The height of the apple at any time t is

 

h(t) = 1/2 g t^2 + v_0 t + h_0, where

g is the acceleration due to gravity (-32 ft/sec/sec),

v_0 is the initial velocity (zero here), and

h_0 is the initial height (16 ft).

 

When the apple hits the ground its height is zero, so

0 = 1/2 * (-32) t^2 + 16

Divide both sides by 16:

0 = -t^2 + 1

t^2 = 1

t = ±1 sec

But since we're only interested in t > 0:

t = 1 sec

Answer 2

Hi L;

The equation you are looking for is...

s=(1/2)at²

s is the distance, which is 16 feet.

a is the acceleration, which, at free-fall, is 9.8 m/s².

t is the time, which is what we are looking for.

 

First, we must convert 16 feet into meters.  I do not know if you are using significant figures, so I will not use such.

16 feet [(0.3048 meters)/(1 foot)]

Let's cancel units...

16 feet [(0.3048 meters)/(1 foot)]

4.8768 meters

 

s=(1/2)at²

(2s)/a=t²

√[(2s)/a]=t

√[(2)(4.8768 m)/(9.8 m/s²)]=t

Let's cancel units...

√[(2)(4.8768 m)/(9.8 m/s²)]=t

√[(2)(4.8768)/(9.8 /s²)]=t

Let's move the //s² from the denominator into the numerator...

√[(2)(4.8768 s²)/(9.8)]=t

Let's calculate...

√(1 s²)=t

1 s=t

 

It will require one second for the apple to reach the ground.

This can be easily visualized.

The distance is 4.8768 meters.

The acceleration is 9.8 m/s².

s=(1/2)at²

4.8768=(1/2)(9.8 m)(t²)

4.8768=(4.9 m)(t²)

 

Answer 3

Hi L -- after 1 second, the speed is 32ft/s ... average speed during that second is 16ft/s ...

 

since distance is V ave * t, a one-second freefall travels the 16ft ... Best regards :)