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<title>A stone is thrown at a velocity v at an angle \varphi to the horizon. After what time will the velocity be at the angle \alpha with the horizon?</title>
$1.3.5.$ A stone is thrown at a velocity $v$ at an angle $\varphi$ to the horizon. After what time will the velocity be at the angle $\alpha$ with the horizon?
</p>
<h3>Solution</h3>
<p>
The horizontal component of velocity remains unchanged:
$$ v_{x}(\varphi) = v_{x}(\alpha) = v \cdot\cos{\varphi}$$
And the horizontal component decreases, depending on time, according to the law:
$$ v_{y}(t) = vt \sin{\varphi} - gt $$
From where the angle that the velocity makes with the horizon is determined as:
$$\tan{\alpha} = \frac{v_{y}(t)}{v_x}$$
Or,
$$\tan{\alpha}\cdot v \cdot\cos{\varphi} = v \cdot\sin{\varphi} - gt $$
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<meta name="description" content="A stone is thrown at a velocity v at an angle \varphi to the horizon. After what time will the velocity be at the angle \alpha with the horizon?">
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<meta property="og:title" content="A stone is thrown at a velocity v at an angle \varphi to the horizon. After what time will the velocity be at the angle \alpha with the horizon?">
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<meta property="og:description" content="A stone is thrown at a velocity v at an angle \varphi to the horizon. After what time will the velocity be at the angle \alpha with the horizon?">
<meta property="og:description" content="A stone is thrown at a velocity v at an angle \varphi to the horizon. After what time will the velocity be at the angle \alpha with the horizon?">
<title>A stone is thrown at a velocity v at an angle \varphi to the horizon. After what time will the velocity be at the angle \alpha with the horizon?</title>
<title>A stone is thrown at a velocity v at an angle \varphi to the horizon. After what time will the velocity be at the angle \alpha with the horizon?</title>
$1.3.5.$ A stone is thrown at a velocity $v$ at an angle $\varphi$ to the horizon. After what time will the velocity be at the angle $\alpha$ with the horizon?
$1.3.5.$ A stone is thrown at a velocity $v$ at an angle $\varphi$ to the horizon. After what time will the velocity be at the angle $\alpha$ with the horizon?
</p>
</p>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
The horizontal component of velocity remains unchanged:
The horizontal component of velocity remains unchanged:
$$ v_{x}(\varphi) = v_{x}(\alpha) = v \cdot\cos{\varphi}$$
$$ v_{x}(\varphi) = v_{x}(\alpha) = v \cdot\cos{\varphi}$$
And the horizontal component decreases, depending on time, according to the law:
And the horizontal component decreases, depending on time, according to the law:
$$ v_{y}(t) = vt \sin{\varphi} - gt $$
$$ v_{y}(t) = vt \sin{\varphi} - gt $$
From where the angle that the velocity makes with the horizon is determined as:
From where the angle that the velocity makes with the horizon is determined as:
$$\tan{\alpha} = \frac{v_{y}(t)}{v_x}$$
$$\tan{\alpha} = \frac{v_{y}(t)}{v_x}$$
Or,
Or,
$$\tan{\alpha}\cdot v \cdot\cos{\varphi} = v \cdot\sin{\varphi} - gt $$
$$\tan{\alpha}\cdot v \cdot\cos{\varphi} = v \cdot\sin{\varphi} - gt $$
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
