• Support PF! Buy your school textbooks, materials and every day products Here!

Projectile Motion Question

  • #1
I am trying to find the initial velocity of a projectile that is launched at a 45 degree angle from a vertical height of 1 meter that lands 2.5 meters in the horizontal direction away at vertical height of 0 meters. Is it possible to calculate without knowing the time?
 

Answers and Replies

  • #2
371
1
yes, use V^2 = V0^2 + 2a([tex]\Delta[/tex]r)

where r is the position
 
  • #3
Thank you that helped me figure it out. Now I have another question. Basically the same as before except the projectile (massless ball) is bounced at some point before reaching its final landing spot 2.5 meters away. The initial angle for this one is -45 degrees. What's the best approach for this?
 
  • #4
I assume that the ball bounces as an elastic collision. Any ideas?
 
  • #5
960
0
I would assume so as well, but you state that the ball has no mass--if so why did it land in the first place? Does it really say that?
 
  • #6
Yeah it says that, may be a typo? I can't really figure this problem out. I've gotten it down to two variable, the initial velocity and the point at which to bounce it.
 
  • #7
960
0
Show me what you have done so far, maybe I can help out? As the question stands, it is absurd. But maybe what it meant to say is point mass, meaning air resistance can be ignored. Because of the elasticity, the angle of incidence and reflection are the same owing to conservation of momentum, (only the ball has no momentum!.) BTW, when you say -45, I assume that is the angle at which the ball is thrown, not at which it strikes the ground.
 
  • #8
Show me what you have done so far, maybe I can help out? As the question stands, it is absurd. But maybe what it meant to say is point mass, meaning air resistance can be ignored. Because of the elasticity, the angle of incidence and reflection are the same owing to conservation of momentum, (only the ball has no momentum!.) BTW, when you say -45, I assume that is the angle at which the ball is thrown, not at which it strikes the ground.
The ball is thrown at -45 from the horizontal from a height of 1 meter. It must bounce and then land 2.5 meters in the x direction at a height of 0.1 meters in the y direction. Find the initial velocity.

I do believe that you're correct about it being a point mass. I know that the angle of incidence and reflection must be the same, however I can't figure out how to get that angle. It seems to me that there are two variables, the initial velocity and the point at which the ball bounces. If I could find where the ball needs to bounce I could solve for the initial velocity. I'm just not sure how to do so.
 
  • #9
960
0
I was stuck too as it seems like it could have been thrown anywhere along the parabolic arc after leaving the hand. The 1.0 meter height helps big time. It lands at 0.1 meters.

Let me throw out a few thoughts about what we do know: the initial Vx=-Vy and the final velocities will equal the initial velocities at the bounce. We also know that the two separate arcs equal a total of 2.5 m and the overall y displacement is 0.9m--anything I am missing?
 
Last edited:
  • #10
It lands at 0 meters when it bounces and then lands at 0.1 meters after the bounce.
 
  • #11
I made a picture to help visualize the problem. Everything you said sounds right except I don't think the final velocity will equal the velocity at the bounce because it doesn't land at the same height.

vmrs41.jpg
 
Last edited:
  • #12
960
0
Well done, except don't you think that the apogee of the arc after the bounce will be higher than the 1.0 meter starting point. It strikes me that at 1.0 meter high after the bounce, the ball will be traveling at Vx upwards and Vx laterally. We don't know the relative displacement at where this occurs, but I think this simplifies the problem but need a few minutes to check on this.
 
  • #13
So are you saying that the angle of the bounce will be 45 degrees?
 
  • #14
960
0
I think this helps, tho can't claim there are not more straightforward methods. I got to wipe the rust off after being gone from PF for so ong, but there should be enough information. Try setting the problem up as you did the first one--you do not know the lateral displacement (call it x) between the point after the bounce where altitude is 1.0 m and its landing point so it can't be solved explicitly. The other part of the flight are two symmetrical arcs (with the velocity sign changed due to the elastic collision) that cause the ball to be displaced by 2.5-x meters. Moreover we know the relationship between the velocity as it leaves the hand and the velocity as it strikes the ground. Use the same eqn as before relating velocity to the displacement of 1.0 meters. Interesting problem.
 
  • #15
960
0
So are you saying that the angle of the bounce will be 45 degrees?
No, not at all. That at the same 1.0 meter height after the bounce the ball will be traveling at +45 degrees angle and Vx=Vy. Think of this as analagous to the first part of the problem, only you have to solve in terms of x, and not explicitly. See above as to where to go from there.
 
  • #16
656
2
I am sort of at a loss as to what is being asked so I will ask if it is possible this is a simple conservation of energy problem?
 
  • #17
I am sort of at a loss as to what is being asked so I will ask if it is possible this is a simple conservation of energy problem?
I'm not sure how to solve it as an energy problem? The picture shows all the given information and the problem is to find the initial velocity.
 
  • #18
960
0
Maybe my crummy drawing will help. PF crummy drawing.jpg
 
  • #19
So how do I find x? Once I know that then it's simply the same as the first problem.
 
  • #20
960
0
I don't think you can find x without using the other piece of the problem, that is the descent and the initial ascent up to the point which is diagrammed. Then you solve simultaneously.

And it may be ugly as both sides of the problem are simultaneous. For an intro class, math shouldn't be that rough--must be something I am not seeing or so dumb math error. Sorry if this has been a wild goose chase,
 
  • #21
I don't think you can find x without using the other piece of the problem, that is the descent and the initial ascent up to the point which is diagrammed. Then you solve simultaneously.

And it may be ugly as both sides of the problem are simultaneous. For an intro class, math shouldn't be that rough--must be something I am not seeing or so dumb math error. Sorry if this has been a wild goose chase,
The two questions are completely independent of each other.
 
  • #22
Anyone have any more ideas for this problem? I'm stuck...
 
  • #23
rl.bhat
Homework Helper
4,433
5
If you throw the projectile horizontally, its total initial energy = 1/2*m*Vo^2 + mgh1
When it lands on the final position, its total energy = 1/2*m*V^2 + mgh2.
Since h1, h2 and R remain constant, the final landing will remain the same for any other velocity whose horizontal component is Vo. The total final energy may change.
Find time t to fall 0.9 m.
Vo = 2.5/t.
Velocity of projection V1 at angle 45 degree is V0*sqrt(2)
 
Last edited:
  • #24
If you throw the projectile horizontally, its total initial energy = 1/2*m*Vo^2 + mgh1
When it lands on the final position, its total energy = 1/2*m*V^2 + mgh2.
Since h1, h2 and R remain constant, the final landing will remain the same for any other velocity whose horizontal component is Vo. The total final energy may change.
Find time t to fall 0.9 m.
Vo = 2.5/t.
Velocity of projection V1 at angle 45 degree is V0*sqrt(2)
By assuming I need to calculate the total time for the ball to fall 1 meter, bounce, come back up and land at 0.1 meters I calculated it to be t=1.332s and the initial horizontal velocity would be V=1.876m/s. To find the initial velocity thrown at 45 degrees it would be V=2.653m/s?
 
Last edited:
  • #25
rl.bhat
Homework Helper
4,433
5
If you through the projectile horizontally, its vertical component is zero. It land on the final spot without bouncing.
The time taken to fall 0.9 m is 0.429s.
Horizontal velocity to cover 2.5 m in 0.429 s is
Vo = 2.5/0.429s = 5.833 m/s.
To land on the same final spot, the other velocities with different angles of projection are such that their horizontal component must be 5.833 and time of flight must be 0.429. Because of the larger velocity it will move larger distance because of bouncing.
In the given problem the angle of projection is 45 degree. So the initial velocity will be such that vi*sin45 = 5.833. Find Vi.
 

Related Threads for: Projectile Motion Question

  • Last Post
Replies
5
Views
916
  • Last Post
Replies
1
Views
870
  • Last Post
Replies
3
Views
762
  • Last Post
Replies
6
Views
897
  • Last Post
Replies
3
Views
786
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
964
  • Last Post
Replies
9
Views
1K
Top