How to find the tangential acceleration in projectile launch?

Jun 2017
337
6
Lima, Peru
The problem is as follows:

A sphere is launched with an initial speed of $50\,\frac{m}{s^2}$ as indicated in the diagram from below. Assume that the acceleration due gravity is $g=10\,\frac{m}{s^2}$ and the angle with the horizontal is $53^{\circ}$. Given these conditions find the tangential acceleration after two seconds produced the launching of the sphere.



This problem doesn't have alternatives.

I'm confused exactly how should I approach this problem:

The only thing which I could spot is that the components would be in the form of:

$a_n=g\cos \omega$

$a_t=g\sin \omega$

where $\omega$ is the launch angle.

But I don't know exactly how to relate this with what would be happening after two seconds from launching?. I really would appreciate an answer which can show where are the vectors for the acceleration and the angles, because I dont know how to identify them properly. Can someone help me with this matter?. Please!!.
 

topsquark

Math Team
May 2013
2,443
1,012
The Astral plane
The diagram for the accelerations will be a tangential acceleration plus a radial (or centripetal) acceleration being equal to the overall acceleration g.

Let the angle between the horizontal and \(\displaystyle \vec{a_t}\) be \(\displaystyle \theta\), which is not equal to \(\displaystyle \omega\) in general. Then, defining the +x direction to the left and +y diretion downward. (I picked these directions simply to make the components positive. There's nothing special about the choice of axes.) Then we have
\(\displaystyle \vec{a_ t} + \vec{a_r} = \vec{g}\).

In components:
\(\displaystyle a_t ~ sin( \theta ) + a_r ~ cos( \theta ) = 0\)
\(\displaystyle a_t ~ cos( \theta ) + a_r ~ sin( \theta ) = g\)

How do you find \(\displaystyle \theta\) at a given time t?

-Dan
 
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skeeter

Math Team
Jul 2011
3,266
1,766
Texas
partial path of the projectile with $\theta$ the direction of $v(2)$ tangent to that path, relative to the horizontal ...

projectile_acc_components.jpg
 
Jun 2017
337
6
Lima, Peru
The diagram for the accelerations will be a tangential acceleration plus a radial (or centripetal) acceleration being equal to the overall acceleration g.

Let the angle between the horizontal and \(\displaystyle \vec{a_t}\) be \(\displaystyle \theta\), which is not equal to \(\displaystyle \omega\) in general. Then, defining the +x direction to the left and +y diretion downward. (I picked these directions simply to make the components positive. There's nothing special about the choice of axes.) Then we have
\(\displaystyle \vec{a_ t} + \vec{a_r} = \vec{g}\).

In components:
\(\displaystyle a_t ~ sin( \theta ) + a_r ~ cos( \theta ) = 0\)
\(\displaystyle a_t ~ cos( \theta ) + a_r ~ sin( \theta ) = g\)

How do you find \(\displaystyle \theta\) at a given time t?

-Dan
Yay! I totally overlooked that the angle wouldn't be the same. But answering your question. I'm not very sure. But I believe this has to do with knowing the direction of the velocity vector at that given instant.

For such thing it would be:

$\left\langle v_o \cos \omega, v_o \sin \omega -gt \right\rangle$

In this given context it would be as follows:

$\vec{v}=\left\langle 50 \cos 53^{\circ}, 50 \sin 53^{\circ} -10t \right\rangle$

which is reduced to:

$\vec{v}=\left\langle 50 \left(\frac{3}{5}\right), 50 \left(\frac{4}{5}\right) -10t \right\rangle$

$\vec{v}=\left\langle 30, 40 -10t \right\rangle$

Then evaluating in $t=2\,s$ it would become into:

$\vec{v}=\left\langle30,20\right\rangle$

But from looking at those lengths it doesn't seem that it will produce a known angle. In the sense of being special angles.

What to do from here?. Can you enlighten me please?
 
Jun 2017
337
6
Lima, Peru
partial path of the projectile with $\theta$ the direction of $v(2)$ tangent to that path, relative to the horizontal ...

View attachment 10892
So I only need to:

$g\sin\theta$

Therefore:

$\sin\theta=\frac{20}{10\sqrt{13}}$

so:

$10\left(\frac{20}{10\sqrt{13}}\right)=5.547\frac{m}{s^2}$

Did I got to the right answer?.