# The minimum speed during the movement of a body thrown at a certain angle to the horizon is 5 m / s

**The minimum speed during the movement of a body thrown at a certain angle to the horizon is 5 m / s, and the maximum speed is 10 m / s. Determine in degrees the angle at which the body is thrown. Neglect air resistance.**

Vmin = 5 m / s.

Vmax = 10 m / s.

∠α -?

The motion of a body can be divided into two types: horizontally, it moves uniformly with a speed Vх, vertically, it is uniformly accelerated with an acceleration of gravity g.

The speed of the body V will be the vector sum of the horizontal component of the velocity Vх and the vertical component Vу: V = Vх + Vу.

Since the velocities Vх, Vу are mutually perpendicular, the resulting velocity V will be the hypotenuse of a right-angled triangle, the legs of which are Vх, Vу.

The minimum speed of the body will be at the top point of the trajectory Vmin, when the body has only a horizontal component Vx: Vmin = Vx.

The body will have the maximum speed Vmax at the moment of throwing the body and at the moment of falling. By the Pythagorean theorem Vmax2 = Vх2 + Vу2.

cosα = Vх / Vmax = Vmin / Vmax = 5 m / s / 10 m / s = 0.5.

According to the table of values of trigonometric functions, the angle ∠α = 60 ° corresponds to cos60 ° = 0.5.

Answer: the body was thrown at an angle ∠α = 60 ° to the horizon.