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

AI Thread Summary
The discussion focuses on determining the velocity and direction of a projectile thrown at a 60-degree angle after specific time intervals. The initial velocity is confirmed to be 10 m/s, and the equations of motion are clarified, including the correct formulas for vertical and horizontal velocities. At half the time to maximum height, the vertical velocity is zero, indicating the projectile is at its peak, while the horizontal velocity remains constant. After the full time of flight, the projectile impacts the ground with a specific horizontal and downward vertical velocity. Clear communication of initial conditions is emphasized for accurate calculations.
Monoxdifly
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Something was thrown with an elevation angle $$60^{\circ}$$. Determine its velocity and direction after:
a. $$\frac{1}{2}\sqrt3$$ second
b. $$\sqrt3$$ second

I tried solving the (a) question by substituting the angle to the equation $$y=v_0\sin\alpha t-\frac{1}{2}gt^2$$ and got 0 (the book uses $$10m/s^2$$ 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 $$t_{\text{x max}}=\frac{2v_0\sin\alpha}{g}$$ I got $$\sqrt3$$ second which means the thing hasn't stopped yet in $$\frac{1}{2}\sqrt3$$ second. Did I make a mistake somewhere?
 
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Monoxdifly said:
Something was thrown with an elevation angle $$60^{\circ}$$. Determine its velocity and direction after:
a. $$\frac{1}{2}\sqrt3$$ second
b. $$\sqrt3$$ second

I tried solving the (a) question by substituting the angle to the equation $$y=v_0\sin\alpha t-\frac{1}{2}gt^2$$ and got 0 (the book uses $$10m/s^2$$ 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 $$t_{\text{x max}}=\frac{2v_0\sin\alpha}{g}$$ I got $$\sqrt3$$ second which means the thing hasn't stopped yet in $$\frac{1}{2}\sqrt3$$ 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 $$v_0$$ is 10 m/s.
 
Monoxdifly said:
After I rechecked the book again, no. The $$v_0$$ 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 $$v_0sin\alpha t-gt$$, and substituting $$t=\frac{1}{2}\sqrt3$$ second gives y = 0.
 
Monoxdifly said:
When I was rechecking the book before, I noticed that the y should be $$v_0sin\alpha t-gt$$, and substituting $$t=\frac{1}{2}\sqrt3$$ 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.
 
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