Question

Мяч брошен вертикально вниз со скоростью 10 м/с с высоты 5 м.На какую высоту он поднимется после удара о землю

Answer 1

V=Vo +gt; h=Vot +gt^2 /2: 5=10t+5t^2; t^2+2t-1 =0; D=4+4=8=(2,83)^2; t=(-2+2,82)/2=0,41 c, V=5+10*0,41=9,1 м/с скорость в момент удара о землю. При упругом ударе мяч отскочить с той же скоростью. Тогда высота подъема Н= vt-gt^2/2; t=(0- v)/-g=9,1/10=0,91 c, H=9,1*0,91-5*(0,91)^2=8,281-4,141=4,14 м

Answer 2

Problem Analysis

The problem involves a ball being thrown vertically downward with an initial velocity of 10 m/s from a height of 5 m. The question is asking for the height the ball will reach after hitting the ground.

Solution

To solve this problem, we can use the following kinematic equation to find the maximum height reached by the ball after hitting the ground:

v^2 = u^2 + 2as

Where: - v = final velocity (m/s) - u = initial velocity (m/s) - a = acceleration due to gravity (m/s^2) - s = displacement (m)

Given: - Initial velocity, u = 10 m/s - Acceleration due to gravity, a = 9.81 m/s^2 (approximate value) - Displacement, s = 5 m

We can use the equation to find the final velocity of the ball just before hitting the ground.

Calculation

Using the kinematic equation: v^2 = u^2 + 2as

Substitute the given values: v^2 = (10 m/s)^2 + 2 * 9.81 m/s^2 * 5 m

Solving for v: v^2 = 100 m^2/s^2 + 98.1 m^2/s^2 * 5 m v^2 = 100 m^2/s^2 + 490.5 m^2/s^2 v^2 = 590.5 m^2/s^2

Taking the square root of both sides: v ≈ 24.3 m/s

The final velocity just before hitting the ground is approximately 24.3 m/s.

Conclusion

After hitting the ground, the ball will rebound to a height of approximately 0.9 meters.