Question

На край крыши дома высотой H с расстояния L от дома школьник хочет забросить мяч. При какой минимальной величине v0 начальной скорости это возможно? Под каким углом a к горизонту следует в этом случае бросить мяч? Ускорение свободного падения g. Сопротивление воздуха пренебрежимо мало.

Answer 1

Как бы то ни было, в определенный момент времени, когда мячик коснется крыши, можно записать два следующих равенства

$v\displaystyle L = v\tau\cos\alpha \\ H = v\tau\sin\alpha-g\tau^2/2\\\\ \tau = L/(v\cos\alpha)\\ H = L\tan\alpha-\frac{gL^2}{2v^2}(1+\tan^2\alpha)\\\\ \frac{gL^2}{2v^2}\tan^2\alpha-L\tan\alpha+H+\frac{gL^2}{2v^2}=0\\\\ D = L^2-\frac{gL^2}{v^2}\left(2H+\frac{gL^2}{v^2}\right)\\\\ \tan\alpha = \frac{v^2}{gL}\left(1\pm\sqrt{1-\frac{g}{v^2}\left(2H+\frac{gL^2}{v^2}\right)}\right)$

Нам надо найти наименьшее v при котором корни вообще могут существовать, то есть решаем предельную ситуацию

$\displaystyle 1-\frac{g}{v^2}\left(2H+\frac{gL^2}{v^2}\right)=0\\\\ \frac{v^2}{g} = 2H+\frac{gL^2}{v^2}\\\\ \frac{v^4}{g} - 2Hv^2-gL^2=0\\\\ v^2 = g(H+\sqrt{H^2+L^2})$

При этом

$\displaystyle \tan\alpha = \frac{v^2}{gL} = \frac{H+\sqrt{H^2+L^2}}{L}$

Answer 2

Minimum initial velocity and launch angle for throwing a ball off the roof

To determine the minimum initial velocity (v0) and launch angle (a) required for a student to throw a ball off the roof of a house with a height (H) from a distance (L) away, we can use the principles of projectile motion.

The key factors to consider are the vertical distance the ball needs to travel (H) and the horizontal distance (L) from the house. We can assume that the acceleration due to gravity (g) and air resistance are negligible.

To find the minimum initial velocity, we need to calculate the time it takes for the ball to reach the ground when thrown at an angle. We can use the following equation for the vertical motion:

H = (v0^2 * sin^2(a)) / (2 * g)

where: - H is the height of the house - v0 is the initial velocity of the ball - a is the launch angle - g is the acceleration due to gravity

To find the time of flight, we can use the equation:

T = (2 * v0 * sin(a)) / g

where: - T is the time of flight

Since we want to find the minimum initial velocity, we can assume that the time of flight is the minimum possible. Therefore, we can substitute the value of T into the equation for H:

H = (v0^2 * sin^2(a)) / (2 * g) = (v0 * sin(a) * T) / 2

Simplifying the equation, we get:

v0 = (2 * g * H) / (sin(a) * T)

Now, let's find the launch angle (a) required. We can use the equation for the horizontal motion:

L = v0 * cos(a) * T

Rearranging the equation, we get:

cos(a) = L / (v0 * T)

Taking the inverse cosine of both sides, we can solve for a:

a = arccos(L / (v0 * T))

Therefore, the minimum initial velocity (v0) and launch angle (a) required for the student to throw the ball off the roof can be calculated using the equations mentioned above.

Please note that the specific values of H, L, and g were not provided in the question. To provide a more accurate answer, please provide the values for these variables.

Let me know if there's anything else I can help you with!