[ASK] Determine its velocity and direction after √(3)/2 seconds

[Monoxdifly]

Something was thrown with an elevation angle \(\displaystyle 60^{\circ}\). Determine its velocity and direction after:
a. \(\displaystyle \frac{1}{2}\sqrt3\) second
b. \(\displaystyle \sqrt3\) second

I tried solving the (a) question by substituting the angle to the equation \(\displaystyle y=v_0\sin\alpha t-\frac{1}{2}gt^2\) and got 0 (the book uses \(\displaystyle 10m/s^2\) as the gravity acceleration), which is impossible since at speed 0 the thing must have been stopped, but when I determined the time when it reached the maximum distance using the formula \(\displaystyle t_{\text{x max}}=\frac{2v_0\sin\alpha}{g}\) I got \(\displaystyle \sqrt3\) second which means the thing hasn't stopped yet in \(\displaystyle \frac{1}{2}\sqrt3\) second. Did I make a mistake somewhere?

 

[lfdahl]

Something was thrown with an elevation angle \(\displaystyle 60^{\circ}\). Determine its velocity and direction after:
a. \(\displaystyle \frac{1}{2}\sqrt3\) second
b. \(\displaystyle \sqrt3\) second

I tried solving the (a) question by substituting the angle to the equation \(\displaystyle y=v_0\sin\alpha t-\frac{1}{2}gt^2\) and got 0 (the book uses \(\displaystyle 10m/s^2\) as the gravity acceleration), which is impossible since at speed 0 the thing must have been stopped, but when I determined the time when it reached the maximum distance using the formula \(\displaystyle t_{\text{x max}}=\frac{2v_0\sin\alpha}{g}\) I got \(\displaystyle \sqrt3\) second which means the thing hasn't stopped yet in \(\displaystyle \frac{1}{2}\sqrt3\) second. Did I make a mistake somewhere?

Hi, Monoxdifly!

Is $v_0 = 5 \frac{m}{s}$? For $y$ to be $0$ at time $t = \frac{\sqrt{3}}{2}s$, it seems you have this $v_0$-value.
Furthermore, the starting position $(x_0,y_0)$ is from "ground zero" - or?
If so, you get the max. height at $t = \frac{v_0\sin \alpha}{g}= \frac{\sqrt{3}}{4}s$. This would make sense, since the total time of flight is $\frac{\sqrt{3}}{2}s$.

 

[Monoxdifly]

After I rechecked the book again, no. The \(\displaystyle v_0\) is 10 m/s.

 

[lfdahl]

After I rechecked the book again, no. The \(\displaystyle v_0\) is 10 m/s.

OK, $v_0 = 10 \frac{m}{s}$. Then, at what time is $y = 0$?

 

[Monoxdifly]

OK, $v_0 = 10 \frac{m}{s}$. Then, at what time is $y = 0$?

When I was rechecking the book before, I noticed that the y should be \(\displaystyle v_0sin\alpha t-gt\), and substituting \(\displaystyle t=\frac{1}{2}\sqrt3\) second gives y = 0.

 

[I like Serena]

When I was rechecking the book before, I noticed that the y should be \(\displaystyle v_0sin\alpha t-gt\), and substituting \(\displaystyle t=\frac{1}{2}\sqrt3\) second gives y = 0.

Shouldn't that be:
$$ v_x = v_0\cos\alpha \\ v_y = v_0\sin\alpha - gt$$
After all, they are asking for velocity and direction aren't they?

 

[Monoxdifly]

Shouldn't that be:
$$ v_x = v_0\cos\alpha \\ v_y = v_0\sin\alpha - gt$$
After all, they are asking for velocity and direction aren't they?

Yes, I miswrote the formula in my first post.

 

[lfdahl]

Yes, I miswrote the formula in my first post.

The kinematic equations can easily be derived (starting point is $x_0 = y_0 = 0$):

Data given:

$g = 10 \frac{m}{s^2}$

$v_0 = 10 \frac{m}{s}$

Elevation angle: $\alpha = 60 ^{\circ}$.

Acceleration:
$$a_x(t) = 0\\a_y(t) = -g = -10 \frac{m}{s^2}$$

Velocity:
\[v_x(t) = v_0\cos \alpha = 5 \frac{m}{s}\\ v_y(x)= v_0\sin \alpha-gt = 5\sqrt{3} - 10t \]

Position:
\[x(t) = v_0\cos \alpha \;t = 5t \\y(t) = v_0\sin \alpha \;t-\frac{1}{2}gt^2 = 5\sqrt{3}t-5t^2\]

So, you get:

(a). $t = \frac{\sqrt{3}}{2}$:
Position:
\[x\left ( t = \frac{\sqrt{3}}{2}s \right ) = 5\frac{\sqrt{3}}{2}m\]
\[y\left (t=\frac{\sqrt{3}}{2}s \right ) = 5\sqrt{3}\frac{\sqrt{3}}{2}-5\left ( \frac{\sqrt{3}}{2} \right )^2= \frac{15}{4}m\]

Velocity:
\[v_x\left ( t = \frac{\sqrt{3}}{2} \right ) = 5 \frac{m}{s}= const. \\v_y\left ( t = \frac{\sqrt{3}}{2} \right ) = 5\sqrt{3}-10\frac{\sqrt{3}}{2} = 0 \frac{m}{s}\]

So at time $t = \frac{\sqrt{3}}{2}$, the $v_y (t) =0$ indicating, that we are at the top point of the parabola.
Hence, the direction of the velocity is along the x-axis.

(b). $t = \sqrt{3}$: (time of impact on the ground)
Position: $x = 5\sqrt{3}m$ and $y = 0 m$.
Velocity: $$v_x = 5 \frac{m}{s}$$ and $$v_y = 5\sqrt{3}-10\sqrt{3}=-5\sqrt{3} \frac{m}{s}$$.

It would be a great help, when you write all the data ($v_0, g, \alpha, x_0, y_0$) from the start.