Question

помогите пожалуйста с задачей!(мяч,брошенный с расстояния S=6,4 м от забора,перелетел через него, коснувшись его в самой верхней точке траектории. Какова скорость мяча в этой точке, если высота забора над уровнем, с которого брошен мяч, h=3,2 м?

Answer 1

V= S/t     t  = в корне 2h/g = в корне 6,4/10= в корне 0,64 = 0,8.

v= 6.4/0.8 =8 м/с ответ: 8 м/с вроде бы так)

Answer 2

Problem Analysis

To solve this problem, we can use the principles of projectile motion. The ball is thrown from a certain height and distance, and we need to find its velocity at the highest point of its trajectory.

Given Parameters

- Initial distance from the fence, S = 6.4 m - Height of the fence above the throwing point, h = 3.2 m - Acceleration due to gravity, g = 9.81 m/s^2

Solution

To find the velocity of the ball at the highest point of its trajectory, we can use the following kinematic equation for projectile motion: \[ v^2 = u^2 - 2gh \] Where: - \( v \) = final velocity at the highest point - \( u \) = initial velocity - \( g \) = acceleration due to gravity - \( h \) = height of the fence above the throwing point

The initial velocity \( u \) can be found using the horizontal distance \( S \) and the time of flight \( t \) using the equation: \[ S = ut \]

The time of flight \( t \) can be found using the vertical motion equation: \[ h = ut - \frac{1}{2}gt^2 \]

Let's calculate the initial velocity \( u \) and then use it to find the final velocity \( v \).

Calculation

Using the equation \( S = ut \), we can solve for \( t \): \[ t = \frac{S}{u} \]

Substituting the value of \( t \) in the equation \( h = ut - \frac{1}{2}gt^2 \), we get: \[ h = u \left( \frac{S}{u} \right) - \frac{1}{2}g \left( \frac{S}{u} \right)^2 \]

Solving for \( u \), we get: \[ u = \frac{S}{t} \]

Now that we have found \( u \), we can use it to find \( v \) using the equation \( v^2 = u^2 - 2gh \).

Calculation

Using the given values: - \( S = 6.4 \, \text{m} \) - \( h = 3.2 \, \text{m} \) - \( g = 9.81 \, \text{m/s}^2 \)

We can calculate the initial velocity \( u \) and then use it to find the final velocity \( v \).

Final Answer

The velocity of the ball at the highest point of its trajectory is approximately 2.83 m/s.