Find Placement Vector of Ball Thrown Ballistically at Angle α from Height H

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Homework Help Overview

The problem involves a ball thrown ballistically at an angle α from a height H, with the goal of determining the placement vector at any moment. The discussion centers around the equations of motion, particularly focusing on the acceleration, velocity, and position vectors.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the acceleration and velocity equations, questioning the formulation of the placement vector. There is confusion regarding the signs and terms in the equations, particularly whether to integrate the sine and cosine functions with respect to time.

Discussion Status

Some participants are exploring different interpretations of the placement vector and questioning the correctness of the provided answer. There is acknowledgment of the constant nature of the angle α, and some guidance has been offered regarding the treatment of sine and cosine in the context of integration.

Contextual Notes

Participants note potential misunderstandings regarding the integration of trigonometric functions and the implications of the angle α being constant. There is also mention of discrepancies in the expected outcomes based on differing interpretations of the equations.

Lenjaku
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Homework Statement



ball is thrown in angle α (balistically) from height H.
what is the placement vector in any moment?

Homework Equations





The Attempt at a Solution



Ok I know acceleration is:
a(t) = (0,−g)

it means the velocity is:
v(t)=(v0 cosα , v0 sinα − gt)

But the answer says the location vector is
r(t) = v0 cosαt,H + v0sinαt-(0.5gt2)/2)

shouldn't be:
r(t) = -v0 sinαt,H + v0cosαt-(0.5gt2)/2) ?

When I do it my way the other parts of the problem turns wrong since my rx got a minus ...then I get equation of the type x2=-t (for instance).

And the answer should come with tanα since my cos and sin are different I get cot...I dun get why it comes wrong since even if I mistook the minus I know for sure cos=sin and sin=>cos I can;t integrate it leaving it as it was can it be there is a problem with the answer?
They didn't say the angle was referring to speed but actually the curve the ball was thrown at... can I integrate without integrating the cos and sin?
 
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I think it is better to change the 'sinαt' into 'tsinα'. This makes it more clear that the angle on which the sin operates is a constant α and does not depend on time.

Note that the integration is with respect to the time t and hence the sin and cos terms are to be considered constant terms and not to be integrated.
 
Lenjaku said:

Homework Statement



ball is thrown in angle α (balistically) from height H.
what is the placement vector in any moment?

Homework Equations





The Attempt at a Solution



Ok I know acceleration is:
a(t) = (0,−g)

it means the velocity is:
v(t)=(v0 cosα , v0 sinα − gt)

But the answer says the location vector is
r(t) = v0 cosαt,H + v0sinαt-(0.5gt2)/2)

shouldn't be:
r(t) = -v0 sinαt,H + v0cosαt-(0.5gt2)/2)[/color] ?
Looks like in that last term you and the answer key divided by 2 once too often
When I do it my way the other parts of the problem turns wrong since my rx got a minus ...then I get equation of the type x2=-t (for instance).

And the answer should come with tanα since my cos and sin are different I get cot...I dun get why it comes wrong since even if I mistook the minus I know for sure cos=sin and sin=>cos I can;t integrate it leaving it as it was can it be there is a problem with the answer?
They didn't say the angle was referring to speed but actually the curve the ball was thrown at... can I integrate without integrating the cos and sin?
the angle alpha is the angle with the horizontal at which the ball is thrown initially. It (and it's sin or cos) is therefore a constant.
 
Thank you very much I completely missed it @.@.

Was running in circles XD

Thanks.
 

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