Gravitational Acceleration and Determining Height

[ZdravkoBG]

Homework Statement

A person standing on a bridge overlooking a highway inadvertently drops an apple over the railing just as the front end of a truck passes directly below the railing. If the vehicle is moving at 55km/h and is 12m long, how far above the truck must the railing be if the apple just misses hitting the rear end of the truck?

Homework Equations

The ones we have learned in class are the general dynamics formulas:
X = X0 + V0t + (1/2)a*t
V = V0 + at
V^2 = V0^2 + 2*a*X

The Attempt at a Solution

I drew myself a little diagram of what is happening to help clarify the problem. After that, I decided to use the distance finding formula since the question is asking about height.

X = X0 + V0t + (1/2)a*t

I set my initial distance (where the apple is dropped) to be 0. The velocity at that point in height is 0 (release). What is left is this:

X = (1/2)a*t

This is where I am getting confused, since there is acceleration of the apple (9.8m/s^2) and acceleration of the truck (horizontal). I also do not know what to use for the t (time) in the equation. Any ideas?

 

[Doc Al]

X = X0 + V0t + (1/2)a*t

That last term should be (1/2)a*t².

This is where I am getting confused, since there is acceleration of the apple (9.8m/s^2) and acceleration of the truck (horizontal). I also do not know what to use for the t (time) in the equation.

Hint: Figure out the time it takes the truck (which is not accelerating) to pass under the bridge.

 

[ZdravkoBG]

Thanks for the hint, I had not realized that the truck had no acceleration.

What I did now is I found how much time is needed for the truck to pass the bridge. I used the displacement formula:

X=X0+V0t+(1/2)*a*t^2

I substituted what I knew:

12=0+55t+(1/2)*0*t^2
t=0.27

I am confused by the units. What is 0.27? Hours?

After this, I substituted the found (t) into the displacement for the apple:

X=0+0+(1/2)*9.8*0.27
X=1.32

Again, I am unsure of the units. Any help with figuring those out?

 

[ypudi]

since the truck is not accelerating we don't need to use the equation of motion.
s= ut+1/at^2

convert 55km/hr into m/s and find out the time for the truck to cross.

s=vxt

 

[ypudi]

the time will be in sec and that is the time taken for the ball to reach the top of the truck.
Now use s= ut+1/at^2 where u=0 and t is the calculated time

 

[Doc Al]

Thanks for the hint, I had not realized that the truck had no acceleration.

What I did now is I found how much time is needed for the truck to pass the bridge. I used the displacement formula:

X=X0+V0t+(1/2)*a*t^2

This is fine.

I substituted what I knew:

12=0+55t+(1/2)*0*t^2
t=0.27

Careful here. Use standard units: Distance measured in meters, time measured in seconds.

As ypudi stated, you must convert 55 km/hr to standard units of m/s.

 

[ZdravkoBG]

Thanks guys, helped a bunch!