# How to find the tangential acceleration in projectile launch?

#### Chemist116

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
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

skeeter

#### skeeter

Math Team
partial path of the projectile with $\theta$ the direction of $v(2)$ tangent to that path, relative to the horizontal ...

#### Chemist116

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?

#### Chemist116

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?.