Question

С какой минимальной по величине скоростью следует бросить вертикально вверх камень, чтобы за первые 4 с длина пути камня была равна 50 м? Точка старта находится достаточно высоко над поверхностью земли. Ускорение свободного падения принять равным 10 м/с^2 . Сопротивление воздуха пренебрежимо мало.

Answer 1

S=Vt - gt²/2
V= (S + gt²/2)/t
V= (50+(10×16)/2)/4
V= 32,5 м/с

Answer 2

Minimum initial velocity to throw a stone vertically upwards to cover a distance of 50m in the first 4s

To determine the minimum initial velocity required to throw a stone vertically upwards so that it covers a distance of 50m in the first 4s, we can use the equations of motion.

The key equation we can use is the equation for displacement in vertical motion:

s = ut + (1/2)gt^2

Where: - s is the displacement (distance covered) in meters - u is the initial velocity in m/s - t is the time in seconds - g is the acceleration due to gravity, which is approximately 10 m/s^2

In this case, we want to find the minimum initial velocity (u) required to cover a distance of 50m in the first 4s. We can rearrange the equation to solve for u:

u = (s - (1/2)gt^2) / t

Substituting the given values: - s = 50m - t = 4s - g = 10 m/s^2

u = (50 - (1/2)(10)(4^2)) / 4

Simplifying the equation:

u = (50 - 80) / 4

u = -30 / 4

u = -7.5 m/s

The minimum initial velocity required to throw the stone vertically upwards so that it covers a distance of 50m in the first 4s is -7.5 m/s.

Please note that the negative sign indicates the direction of the velocity, which is upwards in this case.