# The body is thrown at an initial speed of 23 m / s at an angle of 400 to the horizon.

**The body is thrown at an initial speed of 23 m / s at an angle of 400 to the horizon. Neglecting the air resistance, determine for the moment of time 1.1 s after the start of the movement: the speed of the body, the height of the body’s rise.**

V0 = 23 m / s.

∠α = 40 °.

t = 1.1 s.

g = 9.8 m / s2.

V -?

h -?

The motion of a body can be divided into two types: horizontally, it moves uniformly with a speed V0х, vertically uniformly accelerated with an acceleration of gravity g and an initial speed V0у.

V0х = V0 * cosα.

V0у = V0 * sinα.

The height of the rise h is expressed by the formula: h = V0у * t – g * t ^ 2/2 = V0 * sinα * t – g * t ^ 2/2.

h = 23 m / s * sin40 ° * 1.1 s – 9.8 m / s2 * (1.1 s) ^ 2/2 = 10.3 m.

V = √ (Vx ^ 2 + Vy ^ 2).

Vx = V0x = V0 * cosα.

Vх = 23 m / s * cos40 ° = 17.47 m / s.

Vу = V0у – g * t = V0 * sinα – g * t.

Vу = 23 m / s * sin40 ° – 9.8 m / s2 * 1.1 s = 3.94 m / s.

V = √ (17.47 m / s) ^ 2 + (3.94 m / s) ^ 2) = 17.9 m / s.

Answer: the body will move at a speed of V = 17.9 m / s to a height of h = 10.3 m.