Question

Какую минимальную скорость под углом 30 градусов к горизонту нужно сообщить телу, чтобы перебросить его через стену высотой Н=6 м, если точка бросания находится на высоте h=2м от поверхности Земли и стена удалена от нее на расстояние s=10м?

Answer 1

v*Sin(30)*t - (g* t^2)/2 = H-h
v*Cos(30)*t=10

v*t=10/Cos(30)=2*10/sqrt(3)
v*t*Sin(30)=10/sqrt(3)

(g*t^2)/2=10/sqrt(3)-(H-h)
t=sqrt((2*(10/sqrt(3) -(H-h))/g)
v=(2*10/sqrt(3))/t

Answer 2

Calculation of the Minimum Speed:

To calculate the minimum speed required to throw an object at an angle of 30 degrees to the horizon and clear a wall of height H = 6m, with the throwing point located at a height h = 2m from the ground and the wall at a distance s = 10m, we can use the principles of projectile motion.

The minimum speed required can be found by considering the vertical motion of the object. At the highest point of the trajectory, the object should be at a height equal to the sum of the height of the wall and the height of the throwing point above the ground. Therefore, the vertical displacement of the object should be H + h.

Using the equation for vertical displacement in projectile motion, we have:

H + h = (1/2) * g * t^2 where g is the acceleration due to gravity and t is the time of flight.

To find the time of flight, we can use the equation for horizontal displacement in projectile motion:

s = v * cos(theta) * t where v is the initial velocity of the object and theta is the angle of projection.

Rearranging equation to solve for t, we get:

t = s / (v * cos(theta))

Substituting this value of t into equation we have:

H + h = (1/2) * g * (s / (v * cos(theta)))^2

Simplifying further, we get:

v = sqrt((g * s^2) / (2 * (H + h) * cos^2(theta))) Now we can substitute the given values into equation to calculate the minimum speed required.

Calculation:

Given: - Angle of projection (theta) = 30 degrees - Height of the wall (H) = 6m - Height of the throwing point (h) = 2m - Distance to the wall (s) = 10m

Using the value of the acceleration due to gravity (g) as approximately 9.8 m/s^2, we can substitute the values into equation:

v = sqrt((9.8 * 10^2) / (2 * (6 + 2) * cos^2(30)))

Simplifying further:

v = sqrt(490 / (16 * (3/4)))

v = sqrt(490 / 12)

v ≈ 10.54 m/s

Therefore, the minimum speed required to throw the object at an angle of 30 degrees to the horizon and clear the 6m wall is approximately 10.54 m/s.

Please note that this calculation assumes ideal conditions and neglects air resistance.