# Gravitational Acceleration and Determining Height

## 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
Mentor
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.

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?

since the truck is not accelarating we dont 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

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
Mentor
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.

Thanks guys, helped a bunch!