Question

Из многоэтажки высотой 25 м горизонтально брошен мяч который упал на расстоянии 12 метров от дома с какой скоростью был брошен мяч?помогите решить задачу пж

Answer 1

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Answer 2

Problem Analysis

To find the initial velocity at which the ball was thrown horizontally from a height of 25 meters and landed 12 meters away from the building, we can use the equations of motion. Since the ball was thrown horizontally, the initial vertical velocity is zero. We can use the equation:

d = v*t + (1/2)*a*t^2

where: - d is the horizontal distance traveled by the ball (12 meters), - v is the initial horizontal velocity of the ball (what we need to find), - t is the time taken by the ball to travel the horizontal distance (what we need to find), - a is the horizontal acceleration of the ball (which is zero since there is no horizontal force acting on the ball).

Solution

Let's solve the equation for the initial horizontal velocity (v).

d = v*t

We know that the ball was thrown horizontally, so the initial vertical velocity is zero. Therefore, the time taken to travel the horizontal distance (t) is the same as the time taken to fall from a height of 25 meters.

To find the time taken to fall from a height of 25 meters, we can use the equation:

h = (1/2)*g*t^2

where: - h is the height (25 meters), - g is the acceleration due to gravity (approximately 9.8 m/s^2), - t is the time taken to fall (what we need to find).

Rearranging the equation, we get:

t = sqrt(2*h/g)

Substituting the values, we get:

t = sqrt(2*25/9.8) ≈ 2.55 seconds

Now, we can substitute the value of t into the equation for d to find the initial horizontal velocity (v).

12 = v*2.55

Solving for v, we get:

v ≈ 4.71 m/s

Therefore, the ball was thrown horizontally with an initial velocity of approximately 4.71 m/s.

Answer

The ball was thrown horizontally with an initial velocity of approximately 4.71 m/s.