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

• MHB
• Monoxdifly
In summary, the conversation discussed determining the velocity and direction of an object thrown at an elevation angle of 60^{\circ} after a certain amount of time. The conversation included a mistake in the original calculation and the correction of the formula for velocity and position. The final solution was then provided, including the data given and the calculated values for velocity and position at two different times. It was also suggested to include all data at the beginning of the conversation for clarity.
Monoxdifly
MHB
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?

Monoxdifly said:
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 $\frac{\sqrt{3}}{2}s$.

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

Monoxdifly said:
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 $y = 0$?

lfdahl said:
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.

Monoxdifly said:
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?

I like Serena said:
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.

Monoxdifly said:
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.

What is the equation used to determine velocity and direction after √(3)/2 seconds?

The equation typically used to determine velocity and direction after √(3)/2 seconds is: v = a * t, where v is velocity, a is acceleration, and t is time.

How can the velocity and direction after √(3)/2 seconds be calculated?

The velocity and direction after √(3)/2 seconds can be calculated by using the given equation and plugging in the appropriate values for acceleration and time.

What is the unit of measurement for velocity and direction after √(3)/2 seconds?

The unit of measurement for velocity is typically meters per second (m/s) and the unit for direction is usually degrees (°).

What factors can affect the accuracy of determining velocity and direction after √(3)/2 seconds?

Factors such as external forces, air resistance, and friction can affect the accuracy of determining velocity and direction after √(3)/2 seconds. Additionally, human error in measuring the values for acceleration and time can also impact the accuracy of the calculation.

What other methods can be used to determine velocity and direction after √(3)/2 seconds?

Other methods that can be used to determine velocity and direction after √(3)/2 seconds include using a motion sensor or video analysis software to track the movement of an object and calculate its velocity and direction. Additionally, mathematical models and simulations can also be used to determine these values.

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