Projectile motion question

moonpearl · Dec 8, 2004

Hi, I was wondering if anyone could help me with this question?

A dump truck filled with pumpkins is travelling at 20 m/s when a pumpkin falls off the back and onto
the road, 5 m below. When the pumpkin hits the road, it will be
A) 5 m behind the truck.
B) 20 m behind the truck.
C) 40 m behind the truck.
D) less than 1 m behind the truck.

I used the equation d=(1/2)at^2 to find the time that it took the pumpkin to fall 5m. Then, I assumed that the horizontal velocity of the pumpkin is 20 m/s, so after finding time, I used V=d/t, 20=d/1. I found that the pumpkin should be 20m behind the truck, but the solution to this question is D.

Can anyone tell me why?

Thanks a bunch.

Doc Al · Dec 8, 2004

The horizontal motion is independent of the vertical motion. You could figure out the time it takes the pumpkin to fall to the ground, but who cares?

Both pumpkin and truck are traveling at close to the same horizontal speed: 20 m/s. So, in the one second that it takes for the pumpkin to fall, it does travel 20 m -- with respect to the ground. But so does the truck! They move together.

moonpearl · Dec 8, 2004

but that *isn't* the answer....the answer given is D. I understand why it would be 20...but thats supposed to be wrong. I'm sorry, I'm just really confused.

Doc Al · Dec 8, 2004

moonpearl said:

but that *isn't* the answer....the answer given is D. I understand why it would be 20...but thats supposed to be wrong. I'm sorry, I'm just really confused.

Careful... All the answer choices are described with respect to the truck, not the ground. Note that they all say X meters behind the truck. D is correct, since the pumpkin and truck both move together.

kreil · Dec 8, 2004

Separate it into horizontal and vertical:

Vertical:
vi=0m/s
d=5m
a=g=9.81m/s/s
t=?

d=vit+1/2at^2
d=1/2at^2
5=.5(9.81)t^2
t=1.01s

Use this obtained value for t in the horizontal problem

t=1.01s
vi=20m/s
a=0
d=?

d=vit+.5at^2
d=vit
d=(20)(1.01)=20.2m

EDITED: So this is the distance the object would travel after released. However, if you calculate the distance the truck traveled in 1.01s (=20.2) then you subtract them and this is your answer:

20.2 - 20.2=0m

This is less than 1m, making the answer (D)

P.S. I only showed the work since you said you had obtained 20 already.

moonpearl · Dec 8, 2004

thanks for your replies :) i think i finally get it

VietDao29 · Dec 9, 2004

Hi,
I don't think that kreil's answer is not correct.
If t = 1.01s (not 1s, as the pumpkin takes 1.01s to reach the ground) Then the truck will travel 20.2m .And therefore the pumpkin will land 20.2 - 20.2 = 0 m from the truck.
The reason why the pumpkin lands behind the truck is that that the air slows it down.
And if the truck stops, then the pumpkin will hit the truck, not FALLING BEHIND 20m.
I don't know if my answer is correct. But hope so... :-)
Bye bye,

kreil · Dec 9, 2004

You're right vietdao, thanks for pointing it out (I edited my post).

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Projectile motion question

moonpearl

Doc Al

Staff: Mentor

moonpearl

Doc Al

Staff: Mentor

kreil

moonpearl

VietDao29

kreil

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