Question

стрела выпущена вверх со скоростью 5 м в секунду на какую высоту она поднимется и сколько времени длился полет

Answer 1

$\displaystyle v=5\frac ms\\ \\ h=?m\\ \\ t=?s\\ \\ \\ \\ \\ h=\frac{\Delta v^2}{2\times a}\\ \\ h=\frac{v_1^2-v^2}{2\times a},~v_1=0,~a=g\\ \\ h=\frac{v^2}{2\times g}\\ \\ \\ \\ 0=v-g\times t\\ \\ t=\frac{v}{g}\\ \\ \\ \\ \\ h=\frac{5^2}{2\times 10}=1,25m\\ \\ t=\frac{5}{10}=0,5s$

Answer 2

Calculation of the Height and Duration of Flight

To calculate the height reached by an arrow launched upward with a speed of 5 m/s and the duration of its flight, we can use the laws of motion and the principles of projectile motion.

The height reached by the arrow can be determined using the equation for vertical displacement:

Δy = v₀t - (1/2)gt²

Where: - Δy is the vertical displacement or height reached by the arrow - v₀ is the initial vertical velocity (5 m/s in this case) - t is the time of flight - g is the acceleration due to gravity (approximately 9.8 m/s²)

To find the time of flight, we can use the fact that the arrow reaches its maximum height when its vertical velocity becomes zero. At this point, the arrow starts to fall back down due to the force of gravity.

Let's calculate the height and duration of flight:

1. Calculate the time of flight: - The vertical velocity becomes zero at the maximum height. - Therefore, we can use the equation v = v₀ - gt to find the time it takes for the vertical velocity to become zero. - Rearranging the equation, we have t = v₀/g.

Substituting the given values: - v₀ = 5 m/s - g = 9.8 m/s²

We can calculate the time of flight using the formula: t = 5 m/s / 9.8 m/s²

The time of flight is approximately 0.51 seconds.

2. Calculate the height reached: - Now that we have the time of flight, we can substitute it into the equation for vertical displacement. - Using the equation Δy = v₀t - (1/2)gt², we can calculate the height reached by the arrow.

Substituting the given values: - v₀ = 5 m/s - t = 0.51 seconds - g = 9.8 m/s²

We can calculate the height reached using the formula: Δy = (5 m/s)(0.51 s) - (1/2)(9.8 m/s²)(0.51 s)²

The height reached by the arrow is approximately 1.27 meters.

Therefore, the arrow will reach a height of approximately 1.27 meters and the duration of its flight will be approximately 0.51 seconds.

Conclusion

The arrow, when launched upward with a speed of 5 m/s, will reach a height of approximately 1.27 meters and the duration of its flight will be approximately 0.51 seconds.