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

[Lenjaku]

Homework Statement

ball is thrown in angle α (balistically) from height H.
Waht 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 r_x 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?

 

[grzz]

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.

 

[PhanthomJay]

Homework Statement

ball is thrown in angle α (balistically) from height H.
Waht 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) ?

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 r_x 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.

 

[Lenjaku]

Thank you very much I completely missed it @.@.

Was running in circles XD

Thanks.

 

[Nickie]

I would suggest that the placement vector in any moment can be calculated using the equation r(t) = v0 cosαt,H + v0sinαt-(0.5gt2)/2). This equation takes into account the initial velocity (v0), the angle at which the ball is thrown (α), and the acceleration due to gravity (g).

The equation you suggested, r(t) = -v0 sinαt,H + v0cosαt-(0.5gt2)/2), does not take into account the initial height (H) and would not provide an accurate representation of the ball's placement.

It is important to note that the angle (α) in the equation is not referring to the angle of the ball's trajectory, but rather the angle of the initial velocity vector. This angle is necessary to accurately calculate the placement of the ball as it travels through the air.

In order to integrate the equation, you can use techniques such as substitution and integration by parts to handle the cos and sin terms. It is also important to use the correct limits of integration, which would depend on the time interval being considered.

Overall, I would recommend using the provided equation to calculate the placement vector of the ball thrown ballistically at angle α from height H in any moment, as it takes into account all the relevant variables and provides a more accurate representation of the ball's trajectory.