The body was thrown from a height of 12.7 m on the surface of the earth at a speed of 27.4 m / s at an angle of 45

The body was thrown from a height of 12.7 m on the surface of the earth at a speed of 27.4 m / s at an angle of 45 degrees to the horizon, find the maximum height of the rise above the surface of the earth.

For a body thrown at angles to the horizon at a certain speed (in our case, an angle of 45 °, throwing speed v start = 27.4 m / s), the formula is applied to determine the maximum height –

h max = (v start * sin (α)) ² / 2g (where g is the acceleration of gravity, we will round it up to 10 m / s² for convenience, α is the angle of throw relative to the horizon).

In our case, the body has already been lifted 12.7 m above the ground. The formula is valid for a symmetrical throw trajectory, that is, the start and end points of the trajectory are at the same level (!). We have already set this level with a height of 12.7 meters.

Thus, the maximum body height will be the sum of the initially set height and the height of the throw itself.

Find the maximum throw height using the above formula –

h max = (27.4 m / s * sin (45 °)) ² / 2 * 10 m / s² = 9.38 m.

Therefore, the sought maximum height at which the body turned out to be according to the conditions of our example is 22.08 meters (12.7 m + 9.38 m).