Question

Начальная скорость снаряда выпущенного из пушки вертикально вверх равна 200 м / с в точке максимального подъема снаряд разорвался на два одинаковых осколка которые разлетелись в вертикальных направлениях. Осколок полетевший вниз достиг земли имея скорость 5/3Vo через какое время упадет второй осколок

Answer 1

Примечание U - это Uo, U' - скорость первого осколка. U'=5/3U=333 м/с
Первым делом - преобразование: 
Из закона сохранения энергии мы знаем, что 
mgh=mU^2/2, далее 
hmax=U^2/2g 
Поскольку перед самым разрывом U=0, то справедливо: 
mU1=mU2=0 
m-const, значит 
U1=U2=U
Векторы скоростей двух осколков направлены против друг друга:
mgh+mU^2/2=MU'^2/2
U^2=2gh, следовательно
U=sqrt(U'^2-2gh)=sqrt(U'^2-U^2)
U'=U+gt1 
U'=-U+gt2, откуда следует 
t=(U'-sqrt(U'^2-U^2))/g 
t=(U'+sqrt(U'^2-U^2))/g 
t2-t1=(2/g)*sqrt(U'^2-U^2)=(2/10)*sqrt(333^2-200^2)=53 c
Ответ: Второй осколок упадёт спустя 53 секунды

Answer 2

Calculation of Time for the Second Fragment to Fall

To calculate the time it takes for the second fragment to fall, we need to consider the initial velocity of the projectile, the time it takes for the projectile to reach its maximum height, and the time it takes for the first fragment to fall to the ground.

Given: - Initial velocity of the projectile (Vo) = 200 m/s - The projectile reaches its maximum height at the point of explosion - The first fragment falls to the ground with a velocity of 5/3Vo

To find the time it takes for the projectile to reach its maximum height, we can use the equation for vertical motion:

V = Vo + at

At the maximum height, the velocity (V) will be zero. Assuming the acceleration due to gravity (g) is constant and acting in the downward direction, we can rearrange the equation to solve for time (t):

0 = Vo + gt

Solving for t:

t = -Vo / g

Now, we can find the time it takes for the first fragment to fall to the ground. Since the initial velocity of the first fragment is 5/3Vo, we can use the equation:

V = Vo + gt

Substituting the values:

0 = (5/3Vo) + g * t1

Solving for t1:

t1 = - (5/3Vo) / g

Finally, to find the time it takes for the second fragment to fall, we subtract the time it takes for the first fragment to fall from the total time it takes for the projectile to reach its maximum height:

t2 = t - t1

Let's calculate the values:

Using the given value of Vo = 200 m/s, we can calculate the time it takes for the projectile to reach its maximum height:

t = -Vo / g

Substituting the value of Vo = 200 m/s and assuming the acceleration due to gravity g = 9.8 m/s^2:

t = -200 / 9.8 ≈ -20.41 s

Since time cannot be negative, we take the absolute value:

t ≈ 20.41 s

Next, we can calculate the time it takes for the first fragment to fall:

t1 = - (5/3Vo) / g

Substituting the values of Vo = 200 m/s and g = 9.8 m/s^2:

t1 = - (5/3 * 200) / 9.8 ≈ -34.69 s

Again, taking the absolute value:

t1 ≈ 34.69 s

Finally, we can calculate the time it takes for the second fragment to fall:

t2 = t - t1

Substituting the values of t ≈ 20.41 s and t1 ≈ 34.69 s:

t2 = 20.41 - 34.69 ≈ -14.28 s

Taking the absolute value:

t2 ≈ 14.28 s

Therefore, the second fragment will fall to the ground approximately 14.28 seconds after the first fragment.

Please note that the calculations assume idealized conditions and neglect factors such as air resistance.