Question

Стрела вылетает из лука со скоростью 25 м/с вверх с поверхности Земли. Найти высоту на которой скорость стрелы будет равна 10 м/с

Answer 1

Дано: Vo=25 м/с; V=10 м/с; h - ?

h=(V²-Vo²)/(-2g)

h=(10²-25²)/(-20)=

(10-25)(10+25)/(-20)=

-15*35/(-20)=3*35/4=26,25 м.  Это ответ.

Answer 2

Ответ:

26,25m

Объяснение:

1)V=Vo+at

t=(V-Vo)/a

t=(10-25)/-10

t=1,5c

2)H=Vot+at*t/2

H= 37,5 + (-10 * 2,25)/2=26,25(m)

Answer 3

Finding the height at which the arrow's velocity will be 10 m/s

To find the height at which the arrow's velocity will be 10 m/s, we can use the principles of projectile motion. The initial velocity of the arrow is given as 25 m/s upwards from the surface of the Earth.

Let's assume that the arrow reaches its maximum height at a certain point. At this point, the velocity of the arrow will be momentarily zero before it starts to fall back down. This is because the arrow is subject to the force of gravity, which acts to slow it down until it reaches its highest point.

To find the height at which the arrow's velocity is 10 m/s, we need to determine the maximum height it reaches. At this height, the velocity will be momentarily zero.

We can use the equations of motion to solve for the maximum height. The equation that relates the initial velocity, final velocity, acceleration, and displacement is:

v^2 = u^2 + 2as

Where: - v is the final velocity (which is 0 m/s at the maximum height) - u is the initial velocity (25 m/s upwards) - a is the acceleration due to gravity (approximately -9.8 m/s^2, negative because it acts downwards) - s is the displacement (the height we want to find)

Substituting the given values into the equation, we have:

0^2 = (25)^2 + 2(-9.8)s

Simplifying the equation, we get:

0 = 625 - 19.6s

Rearranging the equation to solve for s, we have:

19.6s = 625

s = 625 / 19.6

Evaluating the expression, we find:

s ≈ 31.89 meters

Therefore, the height at which the arrow's velocity will be 10 m/s is approximately 31.89 meters above the surface of the Earth.

Please note that this calculation assumes idealized conditions and neglects factors such as air resistance.