The body was thrown at a speed of 6 m / s at an angle of 60 ° to the horizon

The body was thrown at a speed of 6 m / s at an angle of 60 ° to the horizon. What is the horizontal projection of its speed before falling?

To find the horizontal projection of the thrown body before falling (this speed is constant throughout the entire trajectory of the body), we will use the formula: Vх = V0х = V0 * cosα.

Variables: V0 – throw speed (V0 = 6 m / s); α – body throw angle (α = 60º).

Let’s calculate: Vх = V0 * cosα = 6 * cos 60º = 3 m / s.

Answer: Before the fall, the horizontal projection of the speed of the thrown body was 3 m / s.