A stone thrown vertically up from the surface of the earth fell to the ground after t = 3 s.

A stone thrown vertically up from the surface of the earth fell to the ground after t = 3 s. at what distance L from the throwing point will the stone fall if it is thrown with the same initial speed at an angle of 45 degrees to the horizon?

First, let’s determine v start – the speed with which the stone was thrown up vertically.

A body thrown upwards loses its initial speed at every moment of time, because gravity acts against it, which slows down its movement, imparting speed to it in the opposite direction with the acceleration of gravity – v down = g * t (where g is the acceleration of free fall, we will round it up to 10 m / s² for convenience).

As long as the speed (v up) of the upward movement exceeds the downward speed, the body moves upward, slowing down all the time. Finally, when the velocities equal (v up = v down), the body stops moving up, “stops” at the maximum height of its rise, in order to start free fall with a known acceleration down. It will return to earth at exactly the same speed with which it was launched from it into the heavens.

This is the law that obeys the movement of the body according to the conditions described above. That is, the movement and the accompanying quantities, one might say, are symmetrical for two equal segments of the body’s path – up and down. And if the stone, in our example, thrown upward, returned after 3 seconds, then it flew to its maximum height in 1.5 seconds and the return journey took him the same 1.5 seconds!

Therefore, v start = g * t = 10 m / s² * 1.5 s = 15 m / s.

For a body thrown at angles to the horizon (in our case, at 45 °), the formula is applied to determine the distance of its flight –

S max = v² start * (sin 2a) / g. Substituting the values ​​already known to us, we get the distance that our stone will fly:

L = (15 m / s) ² * 1 (sin 90 ° = 1) / 10 m / s² = 22.5 meters. This is the length and will be the answer to our example.