A body was thrown from a tower 49 m high in a horizontal position at a speed of 5 m / s.

A body was thrown from a tower 49 m high in a horizontal position at a speed of 5 m / s. At what distance from the tower did it fall and what is its speed at the moment it hit the ground.

Neglecting the force of air resistance, the movement of the body horizontally is uniform, and along the vertical it is uniformly accelerated, with an acceleration g = 9.8 m / s2.

The equation of body motion along the vertical:

h = gt ^ 2/2,

where h is the initial height, h = 49 m; t – fall time, s.

t = (2 h / g) ^ 1/2 = (2 × 49 / 9.8) ^ 1/2 = 3.16 s.

Flight range s:

s = V0 t = 5 × 3.16 = 15.8 m.

Body speed at the moment of impact on the ground:

V1x = V0 = 5 m / s;

V1у = gt = 9.8 × 3.16 = 30.97 m / s;

V1 = (V1x ^ 2 + V1y ^ 2) ^ 1/2 = (5 ^ 2 + 30.97 ^ 2) 1 / ^ 2 = 31.37 m / s.