Help with projectile motion problem

  • Thread starter brightside
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  • #1
I'm having some difficulty with a problem about projectile motion. Just some background about me: I'm just beginning an intro to physics course and I don't have much of a math/science background, so I apologize if I make any errors regarding equations and such.

The problem:

A boy kicks a ball horizontally near the edge of a boardwalk, with an initial speed of 9.0 m/s. A blowing wind gives the ball a constant horizontal acceleration of 12 m/s^2. The ball falls into the water directly under the boy. Ignore the effect of air resistance on the vertical motion of the ball.

A) Determine the height of the boardwalk above water.
B) If the blowing wind reverses direction while maintaining the same strength, where does the ball fall when it is kicked with the same initial speed?

I don't have much byway of an attempted solution for A (I haven't even looked at B yet) because I'm really not even sure where to start, but this is the information I know I have:

ax = 12m/s^2
ay = g = -9.8m/s^2
v0 = 9m/s

I tried to plug it into y = y0 + vy0t + 1/2ayt^2 to firstly find t, but I don't have vy0. In the other questions I've been doing, I've been given the angle in relation to the x axis, which allows me to find vy0 and vx0 using v0cosθ and v0sinθ, respectively. But I don't see how to find it with just the information I have.

So I'm left feeling rather lost regarding this question. Can anyone point me in the right direction?
 

Answers and Replies

  • #2
phinds
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You don't have enough information. Are you sure you have fully stated the problem as it was written?
 
  • #3
gneill
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You don't have enough information. Are you sure you have fully stated the problem as it was written?

There's enough information :smile: It's just provided in a sneaky fashion. Consider where the ball lands in the horizontal direction...
 
  • #4
You don't have enough information. Are you sure you have fully stated the problem as it was written?

Yes, that's the question word for word.


There's enough information :smile: It's just provided in a sneaky fashion. Consider where the ball lands in the horizontal direction...

"The ball falls into the water directly under the boy" - that's just confusing me further :) If it lands directly below him, it would seem like there was no projection. But since there is projection, does that mean the wind is blowing towards him and sends the ball back where it started?
 
  • #5
gneill
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"The ball falls into the water directly under the boy" - that's just confusing me further :) If it lands directly below him, it would seem like there was no projection. But since there is projection, does that mean the wind is blowing towards him and sends the ball back where it started?

Yes, that's it, at least for the horizontal direction. The ball still falls vertically as usual.
 
  • #6
phinds
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There's enough information :smile: It's just provided in a sneaky fashion. Consider where the ball lands in the horizontal direction...

You're right, of course. That slid right by me. Good catch.
 
  • #7
PeterO
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Yes, that's it, at least for the horizontal direction. The ball still falls vertically as usual.

I am sure you mean the ball accelerates vertically in the usual way. The ball actually travels on a curved path (vertically) that happens to end up directly below the boy.
 
  • #8
gneill
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I am sure you mean the ball accelerates vertically in the usual way. The ball actually travels on a curved path (vertically) that happens to end up directly below the boy.
Certainly. I was referring to the horizontal and vertical components separately. Sorry if that wasn't clear.
 

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