Help with Projectiles shot horizontally problem

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In summary: I was just wondering if it was posted already or something...thanks!sure, no...I was just wondering if it was posted already or something...thanks!
  • #1
xxpbdudexx
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I am in an introductory physics class and doing one of these problems, and though it probably isn't too difficult it has stumped me, mainly due to the fact that it asks for vi instead of d or something like that. Anyway, I'm just going to paraphrase the question:

Suppose a basketball player 2.45 m tall throws a basketball at that height and it lands with a speed of 12.0 m/s. What is the ball's initial speed?

The equations you might have to use are simple DVAT's:

df=di+vit+.5at^2
vf=vi+at

These 2 primarily.

So,

d=-2.45
vf=12.0 m/s (I think?)
a=-9.80 m/s^2

From this I got t=.707 s, and I think I should be able to get the vi in the next step or two, but I can't figure out what those steps are. Help please?
 
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  • #2
Hi pbdude :smile: Welcome aboard!

Was the ball thrown horizontally? (I am guessing from the title that it was).

Since this is projectile motion, you should break up the DVAT equations into both 'x' and 'y' directions.

If the ball was indeed thrown horizontally, then when it hits the ground it will hit at an angle, correct? So I do not think that we can assume that 12.0 is vf in just the y-direction as you have in your calculation.

Could you please post the question exactly as it was posed? I think there may be a key word or detail that will make this a lot easier :smile:
 
  • #3
Saladsamurai said:
Hi pbdude :smile: Welcome aboard!

Was the ball thrown horizontally? (I am guessing from the title that it was).

Since this is projectile motion, you should break up the DVAT equations into both 'x' and 'y' directions.

If the ball was indeed thrown horizontally, then when it hits the ground it will hit at an angle, correct? So I do not think that we can assume that 12.0 is vf in just the y-direction as you have in your calculation.

Could you please post the question exactly as it was posed? I think there may be a key word or detail that will make this a lot easier :smile:

I was wondering If I had gotten vf right. It didn't sound quite right to me.

Here's the exact problem:

"Libyan basketball player Suleiman Nashnush was the tallest basketball player ever. His height was 2.45 m. Suppose Nashnush throws a basketball horizontally from a level equal to the top of his head. If the speed of the basketball is 12.0 m/s when it lands, what was the ball's initial speed? (hint: Consider the components of final velocity)"

By the way, my teacher gives us the answers first so we can make sure we did it right, and the answer to this is 9.79 m/s.
 
  • #4
hello, i have a problem just like that, and i didnt want to start a new thread, so can i post it here?

anyways mine's like that but the height is 1.5m(a table) and the velocity(ball rolling off of it) is 5m/s...how do i solve for Vx, Vy, and Dx?
 
  • #5
closertolost said:
hello, i have a problem just like that, and i didnt want to start a new thread, so can i post it here?

anyways mine's like that but the height is 1.5m(a table) and the velocity(ball rolling off of it) is 5m/s...how do i solve for Vx, Vy, and Dx?

First you have to solve for time using -1.5m as Dy (since the ball is falling down) and using Vy as 0 and Ay equaling -9.80. From there, you should be able to find time using Df=Di+Vit+1/2at^2.

After finding the time, use d=rt to find Dx.

I don't know how to find Vx and Vy...
 
  • #6
Okay then :smile: So I was wrong, you paraphrased just fine. The problem is a little challenging because so little info was given, But that's good :wink:

I am a little tired, so I might not last too long, but let's get you thinking about it.

Since we know that it was thrown horizontally, what can you say about the initial velocity in the Y-direction? That is, what is (Vo)y?
 
  • #7
sweet, thanks pbdude :P
 
  • #8
closertolost said:
hello, i have a problem just like that, and i didnt want to start a new thread, so can i post it here?

anyways mine's like that but the height is 1.5m(a table) and the velocity(ball rolling off of it) is 5m/s...how do i solve for Vx, Vy, and Dx?

closertolost: Welcome to PF! Could you actually please start your own thread for this question? It gets a little too crowded in these things and we want to make sure we are addressing everyone's questions appropriately. Thanks!
 
  • #9
Saladsamurai said:
closertolost: Welcome to PF! Could you actually please start your own thread for this question? It gets a little too crowded in these things and we want to make sure we are addressing everyone's questions appropriately. Thanks!

sure, no problem!
 
  • #10
Saladsamurai said:
Okay then :smile: So I was wrong, you paraphrased just fine. The problem is a little challenging because so little info was given, But that's good :wink:

I am a little tired, so I might not last too long, but let's get you thinking about it.

Since we know that it was thrown horizontally, what can you say about the initial velocity in the Y-direction? That is, what is (Vo)y?

It is obviously 0 m/s, since x and y are independent of each other. So the question wants me to find (V0)x.

And I know that frictionless horizontal acceleration is 0 so (Vf)x cannot be 12.0 m/s if (v0)x is 9.79 m/s.
 
  • #11
Right. Okay, so from the first kinematic equation you wrote "df=di+vit+.5at^2" we can establish how long the ball was in the air for by applying it in the y-direction.

Then using your second equation in the y-direction you should be able to find (Vf)y.

Knowing (Vf)y and Vf (given) you should be able to find (Vf)x.

But as you told me already, (Vf)x=(Vo)x due to the no horizontal acceleration condition.
 
  • #12
Saladsamurai said:
Right. Okay, so from the first kinematic equation you wrote "df=di+vit+.5at^2" we can establish how long the ball was in the air for by applying it in the y-direction.

Then using your second equation in the y-direction you should be able to find (Vf)y.

Knowing (Vf)y and Vf (given) you should be able to find (Vf)x.

But as you told me already, (Vf)x=(Vo)x due to the no horizontal acceleration condition.
My main, question, I suppose, is, is the question asking me to solve (V0)y or (V0)x? I'm assuming y. But, I also have already assumed that the V initial for y is zero, so what exactly am I solving for?!
 
  • #13
xxpbdudexx said:
My main, question, I suppose, is, is the question asking me to solve (V0)y or (V0)x? I'm assuming y. But, I also have already assumed that the V initial for y is zero, so what exactly am I solving for?!

You have already told me!

So the question wants me to find (V0)x

So find Voy first and then use that along with the total V (which was given) to find Vox.

Think 'right triangles' :wink:
 
  • #14
OK, so Vf of y is -6.93 m/s. You said I could use that to find Vf, thus Vi. How would I go about doing that?
 
  • #15
Saladsamurai said:
You have already told me!



So find Voy first and then use that along with the total V (which was given) to find Vox.

Think 'right triangles' :wink:

Oh my God! I did it using the Pythagorean theorem! Thank you so much for your help!
 
  • #16
The total final velocity V=12.0 m/s is a vector that is aiming at the ground at some angle, right? So think of this vector as being the 'hypoteneuse' of the right triangle whose 'legs' are Vy and Vx.

Get it?
 
  • #17
xxpbdudexx said:
Oh my God! I did it using the Pythagorean theorem! Thank you so much for your help!

How sweet it is!
 
  • #18
Saladsamurai said:
The total final velocity V=12.0 m/s is a vector that is aiming at the ground at some angle, right? So think of this vector as being the 'hypoteneuse' of the right triangle whose 'legs' are Vy and Vx.

Get it?

Yeah, I got it. :)

Thanks again.
 

Related to Help with Projectiles shot horizontally problem

1. What is a projectile?

A projectile is any object that is thrown, shot, or launched into the air and moves along a curved path due to the force of gravity.

2. What is the problem with projectiles shot horizontally?

The problem with projectiles shot horizontally is that they will eventually fall to the ground due to the force of gravity, unlike projectiles shot at an angle which can travel a longer distance.

3. How can I calculate the distance a projectile shot horizontally will travel?

The distance a projectile shot horizontally will travel can be calculated using the formula d = v0t, where d is the distance, v0 is the initial velocity, and t is the time elapsed.

4. Is air resistance a factor in calculating the motion of a projectile shot horizontally?

Yes, air resistance can affect the motion of a projectile shot horizontally, as it will slow down the projectile and cause it to fall to the ground at a faster rate.

5. How can I improve the accuracy of my calculations for a projectile shot horizontally?

To improve accuracy, it is important to take into account factors such as air resistance, wind speed, and the angle of launch. Using more precise measurements and advanced mathematical models can also help improve accuracy.

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