The body was thrown from a height of 50 meters horizontally at a speed of 20 m / s.

The body was thrown from a height of 50 meters horizontally at a speed of 20 m / s. Determine the flight range (x maximum), flight time, speed after one and a half seconds of flight.

The horizontal movement of the ball is uniform, and the vertical movement is uniformly accelerated, with an acceleration g = 9.8 m / s2 directed vertically downward.

Equation of vertical motion of the ball:

H = g t ^ 2/2,

where H is the initial height of the ball, H = 50 m; t is the time of flight of the ball, s.

Ball flight time: t = (2 H / g) ^ 1/2 = (2 × 50 / 9.8) ^ 1/2 = 3.19 sec.

Ball flight range L:

L = V0 t = 20 × 3.19 = 63.8 m.

After a time t1 = 1.5 s:

V1x = V0 = 20 m / s;

V1у = gt1 = 9.8 × 1.5 = 14.7 m / s;

V1 = (V1x ^ 2 + V1y ^ 2) ^ 1/2 = 24.82 m / s.