$1.3.30^*.$ The projectile leaves the cannon at a velocity $V$ at an angle $\alpha$ to the horizon. What time does the projectile approach the cannon?
### Solution
Using the result obtained in [1.3.6](../1.3.6), at some point in time $t$ we can write the following equations for the velocity of a point:
$$
v_x = v \cos\alpha
$$
$$
v_y = v \sin\alpha - gt
$$
The equation for the displacement of a point along the axes will be:
$$
x = vt \cos\alpha
$$
$$
y = vt \sin\alpha - \frac{gt^2}{2}
$$
At any point $M$, the square of the distance $r^2$ from the origin to this point can be found using the Pythagorean theorem. We are looking for the square so as not to bother with extracting the square root, since we do not need the value of $r$ itself:
$$
L_M = r_M^2 = x_M^2 + y_M^2
$$
$$
L_M = (vt \cos\alpha )^2 + \left(vt \sin\alpha - \frac{gt^2}{2}\right)^2
$$
To determine the domain of decrease of the function $L(t)$, we need to find the values of $t$ for which the derivative $L'(t)$ will be negative. Let's simplify $L(t)$ by opening the brackets and using the basic trigonometric identity, and then find the derivative:
First, we determine the points where the left side becomes zero, and then we find the necessary intervals. We get a quadratic equation with respect to $t$; its solution is trivial and I will not give it. We get two roots, which can be written in one expression:
$$
t \in\left[t_1;\min\left(t_2, \frac{2v\sin\alpha}{g}\right)\right],
$t=(V/g)\sqrt{9\sin^{2}\alpha -8}$ with $\sin\alpha >\sqrt{8/9};$
$ t=0$ with $\sin\alpha <\sqrt{8/9}.$
unchanged lines 34
### Statement
### Statement
$1.3.30^*.$ The projectile leaves the cannon at a velocity $V$ at an angle $\alpha$ to the horizon. What time does the projectile approach the cannon?
$1.3.30^*.$ The projectile leaves the cannon at a velocity $V$ at an angle $\alpha$ to the horizon. What time does the projectile approach the cannon?
### Solution
### Solution
Using the result obtained in [1.3.6](../1.3.6), at some point in time $t$ we can write the following equations for the velocity of a point:
Using the result obtained in [1.3.6](../1.3.6), at some point in time $t$ we can write the following equations for the velocity of a point:
$$
$$
v_x = v \cos\alpha
v_x = v \cos\alpha
$$
$$
$$
$$
v_y = v \sin\alpha - gt
v_y = v \sin\alpha - gt
$$
$$
The equation for the displacement of a point along the axes will be:
The equation for the displacement of a point along the axes will be:
$$
$$
x = vt \cos\alpha
x = vt \cos\alpha
$$
$$
$$
$$
y = vt \sin\alpha - \frac{gt^2}{2}
y = vt \sin\alpha - \frac{gt^2}{2}
$$
$$
At any point $M$, the square of the distance $r^2$ from the origin to this point can be found using the Pythagorean theorem. We are looking for the square so as not to bother with extracting the square root, since we do not need the value of $r$ itself:
At any point $M$, the square of the distance $r^2$ from the origin to this point can be found using the Pythagorean theorem. We are looking for the square so as not to bother with extracting the square root, since we do not need the value of $r$ itself:
$$
$$
L_M = r_M^2 = x_M^2 + y_M^2
L_M = r_M^2 = x_M^2 + y_M^2
$$
$$
$$
$$
L_M = (vt \cos\alpha )^2 + \left(vt \sin\alpha - \frac{gt^2}{2}\right)^2
L_M = (vt \cos\alpha )^2 + \left(vt \sin\alpha - \frac{gt^2}{2}\right)^2
$$
$$
To determine the domain of decrease of the function $L(t)$, we need to find the values of $t$ for which the derivative $L'(t)$ will be negative. Let's simplify $L(t)$ by opening the brackets and using the basic trigonometric identity, and then find the derivative:
To determine the domain of decrease of the function $L(t)$, we need to find the values of $t$ for which the derivative $L'(t)$ will be negative. Let's simplify $L(t)$ by opening the brackets and using the basic trigonometric identity, and then find the derivative:
First, we determine the points where the left side becomes zero, and then we find the necessary intervals. We get a quadratic equation with respect to $t$; its solution is trivial and I will not give it. We get two roots, which can be written in one expression:
First, we determine the points where the left side becomes zero, and then we find the necessary intervals. We get a quadratic equation with respect to $t$; its solution is trivial and I will not give it. We get two roots, which can be written in one expression:
$$
$$
t \in\left[t_1;\min\left(t_2, \frac{2v\sin\alpha}{g}\right)\right],
t \in\left[t_1;\min\left(t_2, \frac{2v\sin\alpha}{g}\right)\right],
