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?

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.3.5.$ 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,
$$\tan{\alpha} \cdot v \cdot \cos{\varphi} = v \cdot \sin{\varphi} - gt$$
From where we obtain the required moment of time:
$$\fbox{$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$$
    </p>


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